True or false X=4 for both 1. 3x+10-(5+3)=14 2. 3x+10(3x-2)=60

Answers

Answer 1
the answer to your question is false because if you substitute 4 for x you'll see that it's false
Answer 2
So, to find if X = 4 for both equations provided, we need to substitute the Xs in the equations and see if they balance out.

For the first equation,

3x+10-(5+3)=14
Every time we see an X, we need to place a 4 in its place.
3(4)+10-(5+3)=14

Using the order of operations, PEMDAS, we start to solve the equation.
3(4)+10-(5+3)=14
12 + 10 - (5+3) = 14
12 + 10 - (8) = 14
12 + 10 - 8 = 14
22 - 8 = 14
14 = 14
TRUE.

For the first equation, x does indeed equal 4.
However, we still need to see if x equals 4 for BOTH of the equations.

3x+10(3x-2)=60
3(4) + 10(3(4)-2) = 60
12 + 100 = 60
112 = 60
This isn't true.

To conclude, the answer is false. X doesn't equal 4 for both of the equations provided.

Hope I could help you out!
If my answer is incorrect, or I provided an answer you were not looking for, please let me know. However, if my answer is explained well and correct, please consider marking my answer as Brainliest!  :)

Have a good one.
God bless!

Related Questions

Find the missing values for the exponential function represented by the table below.

xy
-2 4
-1 6
0 9
1
2

a.
-13.5
-20.25

c.
6
4

b.
-13.5
20.25

d.
13.5
20.25

Answers

d if I'm not mistaking

Answer: c.  

6  

4

Step-by-step explanation:

The exponential function is given by :-

[tex]y=Ab^x[/tex], where A is the initial amount , b is common ratio and x is time period.

We know that in exponential functions, the ratio of the consecutive value of y is same.

From the table ,  [tex]b=\dfrac{6}{4}=\dfrac{3}{2}[/tex]

At x = 0

[tex]9=A(\dfrac{2}{3})^0\\\\\Rightarrow\ A=9[/tex]

At x=1

[tex]y=9(\dfrac{2}{3})^1=6[/tex]

At x=2

[tex]y=9(\dfrac{2}{3})^2=4[/tex]

Hence, the missing values for the exponential function : c.  6  4.

The Pyramid of Giza is one of the largest pyramid structures still standing in Egypt. It is a right pyramid with a square base, a base length of 230 m, and height of 150 m. The area of the base is __________ The volume is ________

Answers

Answer:- The area of base is 52900 square meters. The volume is 2654000 cubic meters.


Explanation:-

Base length of right pyramid (square) a =230m

Height of right pyramid = 150m

Area of base of right pyramid[tex]=a^2=(230)^2=52900\ m^2[/tex]

Volume of right pyramid with square base[tex]=\frac{1}{3}a^2h\\=\frac{1}{3}\times52900\times150=2645000\ m^3[/tex]

Thus, the area of base is 52900 square meters and the volume is 2654000 cubic meters.

The area of the base of the pyramid is 52900 m²

The volume of the square base pyramid of Giza is 2645000 m³

The pyramid is a square base pyramid. Therefore,

volume of a square base pyramid;v =  1 / 3 Bh

where

B = base area

h = height

h = 150 m

Therefore,

area of the base = l²

where

l = length of side

area of the base = 230² = 52900 m²

Volume = 1  / 3 × 52900 × 150

volume = 7935000 / 3

volume = 2645000 m³

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A cold front moved in last weekend. In 8 hours overnight, the temperature outside dropped from 14 degrees to -10. What was the average temperature change for each hour?

I need the answer as quick as I can and I put it at max points!

Answers

Alright, so we have 14 degrees and -10 degrees. The difference between 14 and 0 is 14, and the difference between 0 and -10 is 10. 14+10=24 for total change. For the average over 8 hours, we have 24/8=3 degrees

Answer:


Step-by-step explanation:

The difference between 14 and 0 is 14, and the difference between 0 and -10 is 10. 14+10=24 for total change. For the average over 8 hours, we have 24/8=3 degrees

Use ABC to find the value of sin A. See picture below. Thanks!

Answers

sin is equal 35 divided 37

16x^2=100
How many solutions will there be to the following equation?

Answers

16x^2 = 100
x^2 = 100/16
x^2 = 6.25
x = (+-) sqrt 6.25
x = (+-) 2.5

x = 2.5 and x = - 2.5........2 solutions

Answer:

There are two solutions [tex]x=\frac{5}{2},-\frac{5}{2}[/tex]

Step-by-step explanation:

Given : Equation [tex]16x^2=100[/tex]

To find : How many solutions will there be to the following equation?      

Solution :

Equation [tex]16x^2=100[/tex]

Solve the equation,

Divide by 16 both side,

[tex]\frac{16x^2}{16}=\frac{100}{16}[/tex]

[tex]x^2=\frac{100}{16}[/tex]

Taking root both side,

[tex]x=\sqrt{\frac{100}{16}}[/tex]

[tex]x=\sqrt{\frac{10^2}{4^2}}[/tex]

[tex]x=\pm\frac{10}{4}}[/tex]

[tex]x=\pm\frac{5}{2}}[/tex]

Therefore, There are two solutions [tex]x=\frac{5}{2},-\frac{5}{2}[/tex]

How many sixteenth notes would be needed to have the same duration as 3 quarter notes? Represent this as a fraction

Answers

1 quarter note=2 eighth notes
1 eight note=2 sixteenth notes
so

1 quarter note=4 sixteenth note
so
times 3 both sides
3 quarter notes=12 sixteenth notes

In triangle ABC, angle B = 30°, a = 210, and b = 164. The measure of angle A to the nearest degree is a0

Answers

Ah the Law of Sines! Gotta love it! Set up your equation like this:
a/SinA = b/SinB of which we have all but one of those values.  Our particular equation looks like this then: 210/SinA = 164/Sin30.  Cross multiply to get
210Sin30 = 164SinA.  Divide both sides by 164 to get 210Sin30/164 = SinA.  Do the math on the left and get ,6402 = SinA.  Now use the inverse Sin on your calculator (2nd button then Sin button and then enter .6402) and get an angle of 39.8 degrees.

The length of the third side of the triangle would be 106.5 units approx.

What is a triangle?

A triangle is a two - dimensional figure with three sides and three angles.

The sum of the angles of the triangle is equal to 180 degrees.

∠A + ∠B + ∠C = 180°

Given is that in triangle ABC, angle B = 30°, a = 210, and b = 164.

The cosine formula is given as follows -

c² = a² + b² + 2abcos(α)

c = √(a² + b² + 2abcos(α))

c = √(210)² + (164)² + 2(210)(164)cos(30°)

c = 106.5 units approx.

Therefore, the length of the third side of the triangle would be 106.5 units approx.

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Given: The coordinates of iscosceles trapezoid JKLM are J(-b, c), K(b,c), L(a,0), and M(-a,0).
Prove: The diagonals of an isosceles trapezoid are congruent.
As part of the proof, find the length of KM

A) a2+b2+c2
B) (-a+b)2+c2
C) (a+b)2+c2

Answers

The answer for this question is B. Let me know

Answer with explanation:

It is given that, coordinates of Isosceles trapezoid J K L M are J(-b, c), K(b,c), L(a,0), and M(-a,0).

To Prove: The diagonals of an isosceles trapezoid are congruent.

Proof:

  Distance formula , that is distance between two points in x y plane is given by

       [tex]=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}[/tex]

Where, [tex](x_{1},y_{1}),(x_{2},y_{2})}[/tex] are coordinates of two points in the plane.

Length of Diagonal J L

                  [tex]=\sqrt{(a+b)^2+(0-c)^2}\\\\=\sqrt{(a+b)^2+c^2}[/tex]

Length of Diagonal K M

           [tex]=\sqrt{(a+b)^2+(0-c)^2}\\\\=\sqrt{(a+b)^2+c^2}[/tex]

So, we can see that,

  J L = KM  [tex]=\sqrt{(a+b)^2+(c)^2}[/tex]

Hence,The diagonals of an isosceles trapezoid are congruent.

So ,

  [tex]KM=\sqrt{(a+b)^2+c^2}[/tex]

Option C

(6.3 × 1011) ÷ (7 × 105)

Answers

First multiply within the parantheses
(6.3 * 1011)/(7 * 105)
6369.3/735
8.666 or 8.67

Which Graph correctly represents x+2y≤4?

Answers

It's graph B, remember you can use desmos

Answer with Step-by-step explanation:

We are given an inequality:

x+2y≤4

We have to determine its correct graph

In graph A and graph C (0,4) lies in shaded region but (0,4) does not satisfy the inequality

(since, 0+2×4=8 which is not less than or equal to 4)

Hence, A and C are not the graph of this inequality

In graph D (0,2) does not lie in shaded area but it satisfies the inequality

Hence, D is also not the graph of this inequality

Hence, correct graph of x+2y≤4 is:

Graph B

Kara has 90 lollipops, 36 chocolate bars, and 72 gumballs to put in goody bags for her party. What is the largest number of goody bags that Kara can make so that each goody bag has the same number of lollipops, the same number of chocolate bars, and the same number of gumballs?

Answers

the greatest common factor of 90, 36, and 72 = 18

90/18 = 5 lollipops
36/18 = 2 chocolate bars
72/18 = 4 gumballs

answer : Kara can make 18 goody bags, each containing 5 lollipops, 2 chocolate bars, and 4 gumballs

The largest number of goody bags that Kara can make are [tex]18[/tex] .

What is Highest Common Factor ?

Highest or greatest Common Factor is the largest common factor that all the numbers have in common.

We have,

Number of Lollipops [tex]=90[/tex]

Number of chocolate bars [tex]=36[/tex]

Number of gumballs [tex]=72[/tex]

So,

To find the number of bags;

First find out the Highest Common Factor of [tex]90,36,72[/tex];

[tex]90=2*3*3*5[/tex]

[tex]36=2*2*3*3[/tex]

[tex]72=2*2*2*3*3[/tex]

So, from the factors of all numbers we have,

Highest Common Factor [tex]=18[/tex]

Now,

Lollipops [tex]=\frac{90}{18} =5[/tex]

Chocolate bars [tex]=\frac{36}{18} =2[/tex]

Gumballs [tex]=\frac{72}{18} =4[/tex]

So, the largest number of goody bags that Kara can make are [tex]18[/tex] so that each goody bag has [tex]5[/tex] number of lollipops, [tex]2[/tex] number of chocolate bars, and [tex]4[/tex] number of gumballs.

Hence, we can say that the largest number of goody bags that Kara can make are [tex]18[/tex] .

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how much would $500 invested at 6% interest compounded annually be worth after 4 years

Answers

A = P(1 + r)^t
A = 500(1 + .06)^4
A = $631.24
Remember your formula for compound interest, then just plug the numbers in and calculate. The formula is A=P(1+r)^t where A is the new amount or in this case, your answer. P is the principle you invested. r is the interest rate divided by 100 so you will have a decimal. t is the time in years. This means your formula is A=500(1+.06)^4 Now get out your handy, dandy calculator and you got it!!

Which equation is equivalent to 4s=t+2
a. s=t-2
b. s=4/t+2
c. s=t+2/4
d. s=t+6

Answers

4s=t+2  all of the solutions are in terms of s so divide both sides by 4

s=(t+2)/4

the equivalent equation of the equation 4s=t+2 is s = [tex]\frac{t+2}{4}[/tex] .

What is Equivalent equations?

Equivalent equations are algebraic equations that have identical solutions or roots. Adding or subtracting the same number or expression to both sides of an equation produces an equivalent equation. Multiplying or dividing both sides of an equation by the same non-zero number produces an equivalent equation.

According to the question

The equation  

4s=t+2

Now,

its Equivalent equations  is  :

Dividing equation by 4 both side  

i.e

s = [tex]\frac{t+2}{4}[/tex]

Hence , the equivalent equation of the equation is s = [tex]\frac{t+2}{4}[/tex] .

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which statement is true about angles 1 and 2

Answers

Answer:

Option a) They are adjacent angles

Step-by-step explanation:

Given is a picture of a graph with angles 1 to 6 around it.

Out of these angle 2 and 5 are right angles.

1 and 2 are adjacent to each other.

They are not complementary because 1+2 not equals 90

They are neither supplementary since sum does not equal 180

They cannot be vertical because they are not formed by intersection of two lines.

Hence only option a is right

Option a) They are adjacent angles

Explain how you would use a number line to find the absolute value of –12.

Answers

The absolute value is 12. I don't know how you would use the number line to your advantage but that's because that's not how they taught me in school.
This is kind of hard to explain, but I'm going to try. Feel free to comment questions if what I'm going to say doesn't make sense.

Absolute Value
 # of spaces away from zero
∞ how far away a number is from zero

1). Draw a number line.
2). Label it from -12 to 0.
3). Count on the number line from -12 to 0 and see how many units to the right you have to go to reach zero. The answer is 12, so that is the absolute value.
 |-12| = 12

The scale of a map is 0.5 inch : 20 miles. on the map, the distance between two cities is 1.5 inches. what is that actual distance between the two cities?

Answers

0.5 inch : 20 miles
1.5 inch-?
1.5*20/0.5=60 (miles)
answer:60 miles

Answer:

60 miles.

Step-by-step explanation:

We have been given that the scale map is 0.5 inch : 20 miles. On the map, the distance between two cities is 1.5 inches. We are asked to find the actual distance between the two cities.

We will use proportions to solve for the actual distance between both cities as:

[tex]\frac{\text{Actual distance}}{\text{Map distance}}=\frac{20\text{ miles}}{\text{0.5 inch}}[/tex]

[tex]\frac{\text{Actual distance}}{\text{1.5 inches}}=\frac{20\text{ miles}}{\text{0.5 inch}}[/tex]

[tex]\frac{\text{Actual distance}}{\text{1.5 inches}}*\text{1.5 inches}=\frac{20\text{ miles}}{\text{0.5 inch}}*\text{1.5 inches}[/tex]

[tex]\text{Actual distance}=20\text{ miles}*3[/tex]

[tex]\text{Actual distance}=60\text{ miles}[/tex]

Therefore, the actual distance between both cities is 60 miles.

A turtle is 20 5/6 inches below the surface of a pond. It dives to a depth of 32 1/4 inches. What was the change in the turtles position? Write your answer as a mixed number and show your work.

Answers

alrite, so their change is just their difference, namely 32 1/4 - 20 5/6.

let's change those to "improper" fractions first.

[tex]\bf 20\frac{5}{6}\implies \cfrac{20\cdot 6+5}{6}\implies \cfrac{125}{6} \\\\\\ 32\frac{1}{4}\implies \cfrac{32\cdot 4+1}{4}\implies \cfrac{129}{4}\impliedby \textit{let's use an LCD of \underline{12}}\\\\ -------------------------------\\\\[/tex]

[tex]\bf \cfrac{129}{4}-\cfrac{125}{6}\implies \cfrac{(3\cdot 129)+(2\cdot 125)}{12}\implies \cfrac{387-250}{12}\implies \cfrac{137}{12} \\\\\\ \textit{hmmm 12 goes into 137, 11 times, thus}\quad \boxed{11\frac{5}{12}} \\\\\\ \textit{because}\qquad \cfrac{11\cdot 12+5}{12}\implies \cfrac{132+5}{12}\implies \cfrac{137}{12}[/tex]

A rectangle has a base of 3 inches and a height of 9 inches. If the dimensions are doubled, what will happen to the area of the rectangle?

Answers

check the picture below

how many times is 27 into 108?

Answer:

Area will increase by 4 times

Step-by-step explanation:

Given: A rectangle has a base of 3 inches and a height of 9 inches.  

To find: If the dimensions are doubled, what will happen to the area of the rectangle?

Solution:

It is given that a rectangle has a base of 3 inches and a height of 9 inches.

Now, to find if the dimensions are doubled what will happen to the area, first we need to find the original area.

Original area of rectangle, when base is 3 inches and height 9 inches is

[tex]9\times3=27[/tex] square inches

Now, when dimensions are doubled , the base becomes 6 inches and height becomes 18 inches

So, new area becomes [tex]6\times18=108[/tex] square inches.

Now,

[tex]\frac{\text{new area}}{\text{original area} }=\frac{108}{27}[/tex]

[tex]\implies\frac{\text{new area}}{\text{original area} }=\frac{4}{1}[/tex]

Hence, the area will increase by 4 times.

On monday eliza read her book. on tuesday she read three times as long as she read on monday. on wednesday she read 20 minutes less than tuesday. on thursday she read for 20 minutes which was half as long as she read on wednesday. how many minutes did eliza read over the 4-day period

Answers

monday : x
tuesday : 3x
wednesday : 3x - 20 = 40
thursday: 20

3x - 20 = 40
3x = 40 + 20
3x = 60
x = 60/3
x = 20

monday : 20 minutes
tuesday : 3x....3(20) = 60 minutes
wednesday : 40 minutes
thursday: 20
for a total of : 140 minutes <===

The following is a geometric sequence 5,3,1,-1

Answers

No it is not.  All geometric sequences have a common ratio that is a constant found when you divide any term by the previous one.

3/5 !=1/3

So it is not a geometric sequence.

It IS an arithmetic sequence though, as all arithmetic sequences have a common difference that is a constant found when you find the difference between any term and the term preceding it.

3-5=1-3=-1-1=d=2

So there is a  common difference of 2 so this is an arithmetic sequence.  And all arithmetic sequences can be expressed as:

a(n)=a+d(n-1), a(n)=nth term, a=initial term, d=common difference, n=term number, in this case a=5 and d=-2 so

a(n)=5-2(n-1)  which of course can be simplified...

a(n)=5-2n+2

a(n)=7-2n


For a hypothesis test of H0:p1 − p2 = 0 against the alternative Ha:p1 − p2 ≠ 0, the test statistic is found to be 2.2. Which of the following statements can you make about this finding?

The result is significant at both α = 0.05 and α = 0.01.
The result is significant at α = 0.05 but not at α = 0.01.
The result is significant at α = 0.01 but not at α = 0.05.
The result is not significant at either α = 0.05 or α = 0.01.
The result is inconclusive because we don't know the value of p.

Answers

The test result is not significant at α = 0.05 or α = 0.01.

For a hypothesis test of H0:p1 − p2 = 0 against the alternative Ha:p1 − p2 ≠ 0, the test statistic is found to be 2.2. Given the information provided, the result is not significant at either α = 0.05 or α = 0.01. This conclusion is drawn based on the p-value and the 95% confidence interval.

On compared the test statistic to critical values for α = 0.05 and α = 0.01. Since the test statistic falls within the critical region for α = 0.05 but not for α = 0.01, we concluded that the result is significant at α = 0.05 but not at α = 0.01. The correct options B.

To determine the significance of the test statistic at different levels of significance, we need to compare it to critical values associated with the chosen alpha levels.

Given that the test statistic is 2.2, we need to refer to the critical values of the test statistic for a two-tailed test at α = 0.05 and α = 0.01. These critical values are typically obtained from statistical tables or software.

Let's assume the critical value at α = 0.05 is [tex]\( z_{\alpha/2} = \pm 1.96 \)[/tex]and the critical value at α = 0.01 is [tex]\( z_{\alpha/2} = \pm 2.58 \)[/tex](for a standard normal distribution).

If the test statistic falls within the range defined by these critical values, we can conclude that the result is significant at the corresponding alpha level. Otherwise, the result is not significant.

Since the test statistic of 2.2 falls between the critical values of [tex]\( \pm 1.96 \)[/tex] for α = 0.05 but outside the critical values of [tex]\( \pm 2.58 \)[/tex] for α = 0.01, we can conclude that:

The result is significant at α = 0.05 but not at α = 0.01.

Final answer: The result is significant at α = 0.05 but not at α = 0.01.

We compared the test statistic to critical values for α = 0.05 and α = 0.01. Since the test statistic falls within the critical region for α = 0.05 but not for α = 0.01, we concluded that the result is significant at α = 0.05 but not at α = 0.01. This interpretation aligns with standard hypothesis testing procedures.

Complete question

For a hypothesis test of H0:p1 − p2 = 0 against the alternative Ha:p1 − p2 ≠ 0, the test statistic is found to be 2.2. Which of the following statements can you make about this finding?

A)The result is significant at both α = 0.05 and α = 0.01.

B)The result is significant at α = 0.05 but not at α = 0.01.

C)The result is significant at α = 0.01 but not at α = 0.05.

D)The result is not significant at either α = 0.05 or α = 0.01.

E)The result is inconclusive because we don't know the value of p.

A sample of 4 different calculators is randomly selected from a group containing 18 that are defective and 35 that have no defects. what is the probability that at least one of the calculators is defective?

Answers

1.

P(at least one of the calculators is defective)= 

1- P(none of the selected calculators is defective).

2.

P(none of the selected calculators is defective)

           =n(ways of selecting 4 non-defective calculators)/n(total selections of 4)

3.

selecting 4 non-defective calculators can be done in C(35, 4) many ways, 

where  [tex]C(35, 4)= \frac{35!}{4!31!}= \frac{35*34*33*32*31!}{4!*31!}= \frac{35*34*33*32}{4!}= \frac{35*34*33*32}{4*3*2*1}= 52,360[/tex]

while, the total number selections of 4 out of 18+35=53 calculators can be done in C(53, 4) many ways,

[tex]C(53, 4)= \frac{53!}{4!*49!}= \frac{53*52*51*50}{4*3*2*1}= 292,825[/tex]

4. so, P(none of the selected calculators is defective)=[tex] \frac{52,360}{292,825} =0.18[/tex]


5. P(at least one of the calculators is defective)= 

1- P(none of the selected calculators is defective)=1-0.18=0.82



Answer:0.82

Find the y-intercept and x-intercept of the line.

5x - 4y = 15

y-intercept: __


x-intercept: __

Answers

5x - 4y = 15

x-intercept: y = 0
5x = 15
x = 3
so x-intercept: (3, 0)


y-intercept: x = 0
-4y = 15
y = -15/4
so y-intercept: (0, -15/4)

answer
x-intercept: (3, 0)
y-intercept: (0, -15/4)

PLEASE HELP IMAGE ATTACHED! These triangles are similar. Find the area of the smaller triangle to the nearest whole number.

Answers

Area of a triangle = (1/2)base × height. you have the area and the base so substitute. 105=(1/2)16h... 105/8=13.125 so the height of the first triangle is 13.125.... now set up a proportion using the heights and bases of the similar triangles: 13.125/16=x/12... so x, the height of the second triangle is about 9.84.... now use the original formula to calculate the area of the second triangle... 9.84×12×.5= approx 59 sq ft

The area of smaller triangle is 59 square feet

What are the similar triangles?

Two triangles are similar if they have the same ratio of corresponding sides and equal pair of corresponding angles.

What is the formula for the area of triangle?

The formula for the area of triangle is

[tex]Area = \frac{1}{2} \times base \times \ height[/tex]

According to the given question.

The area of the larger triangle is 105 square feet.

And the one side of the larger triangle and the smaller traingle is 16 and 12 feet respectively.

Suppose the height of thesmaller triangle be x feet and the height of the larger triangle be y feet.

Since, the corresponding edges of similar triangles are proportional.

Therefore,

[tex]\frac{y}{x} = \frac{16}{12}[/tex]

[tex]\implies y = \frac{4}{3} x[/tex]

Also, the area of larger triangle is 105 square feet.

[tex]\implies \frac{1}{2} \times \frac{4}{3} x \times 16 = 105[/tex]

The above euqtaion can be written as

[tex]\implies \frac{1}{2}\times \frac{4}{3} x \times \frac{4}{3} \times 12 = 105[/tex]

[tex]\implies \frac{1}{2} \times (\frac{4}{3} )^{2} \times x \times 12 = 105[/tex]

[tex]\implies \frac{1}{2} \times x \times 12 = 105 \times \frac{9}{16}[/tex]

[tex]\implies \frac{1}{2} \times x \times 12 = 59[/tex]

Area of smaller triangle  = 59 square feet

Hence, the area of smaller triangle is 59 square feet.

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can someone help me please

Answers

The correct answer is C) Graph A. We can find this remembering our rules about graphing lines such as parabolas. You could also use a graphing tool, such as Desmos, to input the equation and see the outcome. I would also recommend purchasing a graphing calculator (TI-84 Plus) which would let you put this equation in and show you the line.

Find all values of $x$ such that $6= \dfrac{35}{x} -\dfrac{49}{x^2}$. If you find more than one value, then list your solutions in increasing order, separated by commas.

Answers

x=7/3 and x=7/2

if you want the step-by-step then you just try to isolate x and it is pretty self-explanatory from there on. Hope this helped! 

:P

Answer:

[tex]x=\frac{7}{3} , \frac{7}{2}[/tex]

Step-by-step explanation:

[tex]6= \frac{35}{x} - \frac{49}{x^2}[/tex]

Now we need to solve for x

To get 'x' alone we make the denominators same

LCD = x^2

WE multiply the whole equation by x^2

[tex]6x^2 = 35x - 49[/tex]

Now we the equation =0, move all the terms to left hand side

[tex]6x^2-35x + 49=0[/tex]

Now we apply quadratic formula to solve for x

a= 6, b= -35 , c= 49

[tex]x= \frac{-b+-\sqrt{b^2-4ac}}{2a}[/tex]

[tex]x= \frac{-(-35)+-\sqrt{(-35)^2-4(6)(49)}}{2*6}[/tex]

[tex]x= \frac{35+-\sqrt{49}}{12}[/tex]

[tex]x= \frac{35+-7}{12}[/tex]

Now frame two equations , one with +  and another with -

[tex]x= \frac{35+7}{12}[/tex]                                  [tex]x= \frac{35-7}{12}[/tex]

[tex]x= \frac{42}{12}[/tex]                                       [tex]x= \frac{28}{12}[/tex]  

[tex]x= \frac{7}{2}[/tex]                                           [tex]x= \frac{7}{3}[/tex]  

So value of x= {7/3, 7/2}

Which of the following equations have graphs that are parallel to the graph of the equation y=-3/2x+8?

I. 3x + 2y = 10
II. 2x − 3y = 9
III. 6x + 4y = 28
IV. 3x − 2y = 8

 I and III only
 II and III only
 IV only
 III only

Answers

parallel lines will have the same slope, but different y int
y = -3/2x + 8....slope = -3/2....y int = 8

(I) 3x + 2y = 10
    2y = -3x + 10
    y = -3/2x + 5....slope = -3/2, y int = 5....this IS parallel

(II) 2x - 3y = 9
     -3y = -2x + 9
     y = 2/3x - 3...slope = 2/3, y int = -3....is not parallel

(III) 6x + 4y = 28
       4y = -6x + 28
       y = -3/2x + 7...slope = -3/2, y int = 7....this IS parallel

(IV) 3x - 2y = 8
      -2y = -3x + 8
      y = 3/2x - 4...slope = 3/2...y int = -4...this is not parallel

solution is : I and III

Math help needed. Thank you

Answers

I think the answer is c
The answer is A. :) Think about it if you have one point and u put a bunch of points around it all the same distance away and u connect the dots u make a circle.

∠EFG and ∠GFH are a linear​ pair, m∠EFG=3n+19, and m∠GFH=55+33 What are m∠EFG and m∠​GFH?

Answers

a linear pair is a pair of angles that form a straight line and are therefore, equal to 180 degrees

3n + 19 + 55 + 33 = 180
3n + 107 = 180
3n = 180 - 107
3n = 73
n = 73/3 


so < EFG = 3n + 19 = 3(73/3) + 19 = 73 + 19 = 92
and < GFH = 55 + 33 = 88

What is the correct name for this circle?

Answers

circle C
hope this helped!
The circle takes name from its center. I'd say Circle C
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