use the concept of slope to find t such that three points are collinear
(-3,3) (t,-1) (8,6)
Express each fraction as a percent .round to the nearest whole number 178 over 450
What other information is needed to prove that the two triangles congruent by SAS?
Picture description: there are two triangles showing line LT = line MQ and L=M
A.
B.
C. Line GT = line NQ <<
D. Line LG = line MN
What other information is needed to prove the two triangles congruent by SAS? Pick 2.
A. S = U<<
B. T = V
C. S = V
D. T = U
E. Line RS = line WU<<<
F. line RT = line VU
Answer:
D. LG = MN
Step-by-step explanation:
Final answer:
To prove two triangles congruent by SAS, we require equality of two sides and the included angle. Given line LT = line MQ and angle L = angle M, we also need line GT = line NQ and line LG = line MN. For the second part, line RS = line WU and line RT = line VU are necessary to fulfill SAS criteria.(Options C and D for first and Option E and F for second)
Explanation:
When proving two triangles are congruent using the Side-Angle-Side (SAS) postulate, you need to know that two sides and the included angle (the angle between the two sides) of one triangle are exactly equal to two sides and the included angle of another triangle.
In the given problem, we're told that line LT is congruent to line MQ and angle L is congruent to angle M. The additional information needed to prove triangle congruence by SAS would be to show that the second pair of sides around the included angle are also congruent (as in option C Line GT = line NQ) and to ensure congruence on the corresponding parts of the other triangle (as in option D Line LG = line MN).
Answering the second part, to prove congruence by SAS, we would need two corresponding sides and the included angle. Therefore, the correct choices are option E Line RS = line WU which provides the second side, and option F line RT = line VU ensures the sides are corresponding in the two triangles, enclosing the angle which is already given as equal.
how many numbers are between 100 and 200?
Point G is the centroid of triangle abc use the information to find the value of x.
1. GC=3x +7 and CE= 6x
2. FG= x +8 and AF = 9x - 6
3. Bg=5x -1 and DG = 4x -5
The value of x is 7/6.
To find the value of x, we can utilize the given information about the lengths of the medians and segments within triangle ABC.
Using medians CE and GC:
Since G is the centroid of triangle ABC, it divides each median into two segments in a ratio of 2:1. Therefore, we can set up two equations based on the given lengths:
CE = 2/3 * GC
6x = 2/3 * (3x + 7)
Solving for x, we get:
6x = 2x + 14/3
4x = 14/3
x = 7/6
Using medians BG and FG:
Similarly, we can set up two equations based on the given lengths of medians BG and FG:
FG = 2/3 * BG
x + 8 = 2/3 * (5x - 1)
Solving for x, we get:
x + 8 = 10x/3 - 2/3
8/3 = 9x/3
x = 8/9
Using medians DG and AG:
Following the same approach, we can set up two equations based on the given lengths of medians DG and AG:
AG = 2/3 * DG
9x - 6 = 2/3 * (4x - 5)
Solving for x, we get:
9x - 6 = 8x/3 - 10/3
x - 6 = 8x/3 - 10/3
-5 = 5x/3
x = -3
Comparing the values obtained from each set of equations, we find that x = 7/6 is consistent across all three sets. Therefore, the value of x is 7/6.
Which of the sets of ordered pairs represents a function?
A = {(−4, 5), (1, −1), (2, −2), (2, 3)}
B = {(2, 2), (3, −2), (9, 3), (9, −3)}
1. Only A
2. Only B
3. Both A and B
4. Neither A nor B
A scale drawing for a restaurant is shown below.
In the drawing, 5cm represents 6m.
Assuming the dining hall is rectangular, find the area of the real dining hall.
What is the area?
The area of the real rectangular dining hall (assuming that 5cm on paper is 6m in real) is 72m².
A rectangle is a quadrilateral whose opposite edges are of equal length and the opposite angles are of equal measurements.
Given that
the length of the dining hall, l = 10cm
the width of the dining hall, b = 5cm
As evident from the figure, the dining hall is rectangular, and since it is known that the area of a rectangle is the product of its length and width, therefore,
Area of the hall (on paper) = l × b
= 10 × 5
= 50 square cm.
Also, as it is given in the question that 5cm on paper represents 6m in real, therefore
[tex]Area\ of\ the\ hall\ (in\ real)= (\dfrac{10}{5}\times6)\times(\dfrac{5}{5}\times6)[/tex]
[tex]\\= 2\times6\times6\\= 72\ m^2[/tex]
Hence, the area of the hall in real is 72m².
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Multiplying a trinomial by a trinomial follows the same steps as multiplying a binomial by a trinomial. Determine the degree and maximum possible number of terms for the product of these trinomials: (x2 + x + 2)(x2 – 2x + 3). Explain how you arrived at your answer.
The product of the given trinomials will have a degree of 4, and since there are 9 terms, the maximum possible number of terms for the product is 9.egree of 4 and a maximum possible number of terms of 9.
When multiplying two polynomials, the degree of the resulting polynomial is the sum of the degrees of the two polynomials being multiplied.
In this case, both trinomials have a degree of 2.
The maximum possible number of terms in the product of two polynomials is the product of the number of terms in each polynomial.
For the first trinomial, there are 3 terms, and for the second trinomial, there are also 3 terms.
Multiplying these together gives a maximum possible number of terms of 9.
To arrive at this conclusion, you can use the distributive property, where each term in the first trinomial is multiplied by each term in the second trinomial, resulting in a total of 9 possible term combinations.
Therefore, the product of the two trinomials will have a dTo multiply two trinomials, like [tex]\( (x^2 + x + 2)(x^2 - 2x + 3) \)[/tex], you apply the distributive property repeatedly, multiplying each term in the first trinomial by each term in the second trinomial and then summing up the products.
Each term in the first trinomial needs to be multiplied by each term in the second trinomial, resulting in a total of [tex]\( 3 \times 3 = 9 \)[/tex] products.
Now, regarding the degree, when you multiply two polynomials, you add the degrees of the polynomials.
Here, both polynomials are of degree 2, so the resulting polynomial will be of degree 2 + 2 = 4.
Thus, the product of the given trinomials will have a degree of 4, and since there are 9 terms, the maximum possible number of terms for the product is 9.egree of 4 and a maximum possible number of terms of 9.
If n is a positive integer, then lim (n-->infinity) (1/n) [1/(1+(1/n)) + 1/(1+(2/n))+...+1/(1+(n/n)] is?
the above can be expressed as
a.) integral from 0 to 1 of (1/x) dx
b.) integral from 1 to 2 of (1/(x+1))dx
c.) integral from 1 to 2 of (x)dx
d.) integral from 1 to 2 of (2/(x+1))
e.) integral from 1 to 2 of (1/x)
The given sequence is in the form of a Riemann sum for the integral of the function 1/(1+x) from 1 to 2. So, the correct answer is b) 'integral from 1 to 2 of (1/(x+1)) dx'.
Explanation:The sequence in the question seems to be in the form of a Riemann sum for an integral. The term inside the summation loop can be expressed as 1/(1+i/n) where i varies from 1 to n. The limit of the sequence as n tends to infinity can be thus represented as the integral from 0 to 1 of the function 1/(1+x) dx. The function 1/(1+x) is a continuous function over the interval [0,1], so the integral exists and the limit of the sequence is legitimately defined.
Therefore, looking in the options provided, the answer to your question will be 'b.) integral from 1 to 2 of (1/(x+1)) dx'. This integral is the limit of the given series as n approaches infinity.
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The limit as 'n' approaches infinity of this summation is equal to the integral from 1 to 2 of (1/(x+1))dx. Essentially, this question is turning the sum into an integral as 'n' approaches infinity.
Explanation:This question is asking for the limit of a sum as n approaches infinity, which is essentially the definition of an integral. Every term in the sum could be expressed as 1/(1+i/n), where 'i' is the sum index. Each term, when 'n' becomes infinitely large, becomes a term of the form i/n, or Δx, which is a small piece of the variable we’re integrating. Similarly, each 1/(1+i/n) term turns into 1/x for that small piece.
Given these transformations and the fact that the index i ranges from 1 to n, it's clear that as n approaches infinity, we are integrating the function 1/x from 1 to 2.
Thus, the correct answer would be: b.) integral from 1 to 2 of (1/(x+1))dx
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Jordan keeps track of rainy days for 1 year . This year he counted 52 weeks in a year with at least I rainy day. How many weeks had no rainy days.
Answer:
A.
Step-by-step explanation:
In a family with eight children, what is the probability that exactly six are boys? a. 7168 b. 28 c. 0.109375 d. 0.015625
The probability that exactly six children in the family are boys is 0.015625.
What is Combination?An arrangement of objects where the order in which the objects are selected does not matter.
The probability of having exactly six boys in a family with eight children can be calculated using the binomial distribution formula:
[tex]P(X = k) =^nC_{k} p^k(1-p)^(^n^-^k^)[/tex]
P(X = k) is the probability of having exactly k boys, n is the total number of children, k is the number of boys we want to find , p is the probability of having a boy and (1-p) is the probability of having a girl (also 1/2 in this case)
Substituting the values into the formula, we get:
P(X = 6) = ⁸C₆ (1/2)⁶ (1/2)⁸⁻⁶
= (8! / (6! ×2!)) (1/2)⁸
= (87/21) × (1/2)^8
= 0.015625
Therefore, the probability that exactly six children in the family are boys is 0.015625.
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factored form of 25x^2+40x+16
Factor each polynomial (1)64-40ab
1. Julia deposited $250 in a savings account that earns 2.7% simple interest. How much interest has Julia earned by the end of the first year?
$6.75
$92.59
$256.75
$675.00
2. Armand deposited $389.42 in a savings account that earns 3.2% simple interest. What is Armand’s account balance after seven years?
$87.23
$401.88
$476.65
$872.30
3. Andy deposited $1,567.12 in a savings account that earns 1.9% simple interest. What will Andy’s account balance be in nine months?
$1,567.12
$1,589.45
$1,596.90
$1,835.10
Answer:
Step-by-step explanation:
Simple interest formula is : [tex]p\times r\times t[/tex]
t is always in years here.
1.
p = 250
r = 2.7% or 0.027
t = 1
Simple interest earned = [tex]250\times0.027\times1[/tex] = $6.75
2.
p = 389.42
r = 3.2% or 0.032
t = 7
Simple interest earned = [tex]389.42\times0.032\times7[/tex] = $87.23
Amount after 7 years will be = [tex]389.42+87.23[/tex] = $476.65
3.
p = 1567.12
r = 1.9% or 0.019
t = [tex]9/12=0.75[/tex]
Simple interest earned = [tex]1567.12\times0.019\times0.75[/tex] = $22.33
Account balance after 9 months = [tex]1567.12+22.33[/tex] = $1589.45
Find all real values of x, such that f(x)=0.
f(x)=15-3x ...?
Find all points of extrema on the interval [0,2pi] if y=x-cosx ...?
The function y=x-cosx has an extremum on the interval [0,2pi] at x=3pi/2, determined by setting the first derivative equal to zero. The second derivative test is inconclusive at this point, but the point (3pi/2, 3pi/2) is a minimum due to the nature of the function.
To find all points of extrema on the interval [0,2pi] for the function y=x-cosx, we first need to find the derivative of the function and determine where it is equal to zero. The first derivative of y=x-cosx is y' = 1 + sinx. The critical points occur where y' = 0, which translates to 1 + sinx = 0 or sinx = -1. The only point in the interval [0,2pi] where sinx = -1 is at x = 3pi/2. To determine if this critical point is a maximum or minimum, we apply the second derivative test.
The second derivative of the function is y'' = cosx. At x = 3pi/2, the second derivative y'' = cos(3pi/2) = 0; since the second derivative is zero, the second derivative test is inconclusive. However, we can analyze the function around the critical point to conclude or use other methods like analyzing the first derivative's sign change.
For y=x-cosx, since the original function is a combination of a linear increasing function and cosine function, the critical point at x = 3pi/2 will be a minimum as the function increases both before and after this point.
Therefore, the point of extrema on the interval [0,2pi] for the function y=x-cosx are at (3pi/2, 3pi/2 - cos(3pi/2)), which simplifies to (3pi/2, 3pi/2).
The half-life of a radioactive substance is the time required for half of a sample to undergo radioactive decay, or for the quantity to fall to half its original amount. Carbon 14 has a half-life of 5,730 years. Suppose given samples of carbon 14 weigh (fraction 5/8) of a pound and (fraction 7/8) of a pound. What was the total weight of the samples 11460 years ago
(show work please)
Answer:
Amount of C-14 taken were 2.5 pounds and 3.5 pounds respectively.
Step-by-step explanation:
Radioactive decay is an exponential process represented by
[tex]A_{t}=A_{0}e^{-kt}[/tex]
where [tex]A_{t}[/tex] = Amount of the radioactive element after t years
[tex]A_{0}[/tex] = Initial amount
k = Decay constant
t = time in years
Half life period of Carbon-14 is 5730 years.
[tex]\frac{A_{0} }{2}=A_{0}e^{-5730k}[/tex]
[tex]\frac{1}{2}=e^{-5730k}[/tex]
Now we take ln (Natural log) on both the sides
[tex]ln(\frac{1}{2})=ln[e^{-5730k}][/tex]
-ln(2) = -5730kln(e)
0.69315 = 5730k
[tex]k=\frac{0.69315}{5730}[/tex]
[tex]k=1.21\times 10^{-4}[/tex]
Now we have to calculate the weight of samples of C-14 taken for the remaining quantities [tex]\frac{5}{8}[/tex] and [tex]\frac{7}{8}[/tex] of a pound.
[tex]\frac{5}{8}=A_{0}e^{(-1.21\times 10^{-4}\times 11460)}[/tex]
[tex]\frac{5}{8}=A_{0}e^{(-1.21\times 10^{-4}\times 11460)}[/tex]
[tex]\frac{5}{8}=A_{0}e^{(-1.3863)}[/tex]
[tex]A_{0}=\frac{5}{8}\times e^{1.3863}[/tex]
[tex]A_{0}=\frac{5}{8}\times 4[/tex]
[tex]A_{0}=\frac{5}{2}[/tex]
[tex]A_{0}=2.5[/tex] pounds
Similarly for [tex]\frac{7}{8}[/tex] pounds
[tex]\frac{7}{8}=A_{0}e^{(-1.21\times 10^{-4}\times 11460)}[/tex]
[tex]\frac{7}{8}=A_{0}e^{(-1.21\times 10^{-4}\times 11460)}[/tex]
[tex]\frac{7}{8}=A_{0}e^{(-1.3863)}[/tex]
[tex]A_{0}=\frac{7}{8}\times e^{(1.3863)}[/tex]
[tex]A_{0}=\frac{7}{8}\times 4[/tex]
[tex]A_{0}=\frac{7}{2}[/tex]
[tex]A_{0}=3.5[/tex] pounds
1.5x-2<10 how do you solve this
Understanding this proof for the proposition "For all integers a, gcd(9a+4, 2a+1) = 1.
Proof: gcd(9a+4, 2a+1) = gcd(2a+1, a) = gcd(a, 1). Since gcd(a, 1)=1, gcd(9a+4, 2a+1) =1.
First line because 4(2a+1)=8a+4 and 9a+4-(8a+4)= a
Second line because a times 2 =2a and 2a+1-2a=1
Although the second equality is more or less obvious since 2a+1 leaves a remainder of 1 when divided by a.
Show that the equation represents a circle by rewriting it in standard form, and find the center and radius of the circle.
5x^2 + 5y^2 + 10x − y = 0
The equation 5x^2 + 5y^2 + 10x - y = 0 represents a circle. By completing the square, the equation is rewritten in the standard form as (x + 1)^2 + (y - 1/10)^2 = 101/100. The circle's center is at (-1, 1/10) with a radius of approximately 1.005.
Explanation:To show that the equation 5x^2 + 5y^2 + 10x - y = 0 represents a circle and to rewrite it in standard form, we can complete the square for both x and y terms. First, factor out the coefficients of the squared terms:
5(x^2 + 2x) + 5(y^2 - 1/5y) = 0Divide both sides by 5 to simplify the equation:
(x^2 + 2x) + (y^2 - 1/5y) = 0Now, complete the square by adding and subtracting the necessary constants inside each parenthesis:
(x^2 + 2x + 1) - 1 + (y^2 - 1/5y + 1/100) - 1/100 = 0By completing the square, the equation becomes:
(x + 1)^2 + (y - 1/10)^2 = 1 + 1/100This can be further simplified to:
(x + 1)^2 + (y - 1/10)^2 = 101/100The standard form of the equation for a circle is (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center and r is the radius of the circle. Therefore, our circle's center is at (-1, 1/10) and its radius is the square root of 101/100, which simplifies to 10.05/10 or approximately 1.005.
The distance from the school to Brandi's house is 1,240 meters. Leaving the school, she rides her bicycle for 60 seconds at a speed of 5 meters per second. If Brandi continues cycling at this speed, how many more seconds will it take her to arrive at her house?
Brandi rides her bike at a speed of 5 m/s from school to her house. She will take an additional 188 seconds to cover the remaining distance of 940 meters to arrive at her house.
Explanation:Brandi's situation:
Distance to house: 1,240 metersSpeed leaving school: 5 m/sTime taken initially: 60 secondsTo find: Additional time needed to reach house.
Calculations:
Time taken initially: 60 secondsRemaining distance: 1,240 - (60*5) = 940 metersTime to cover the remaining distance: 940 / 5 = 188 secondsTotal time: 60 + 188 = 248 secondsChoose the equation of the vertical line passing through the point (-4, 2). i picked x = -4.. ...?
Which type of triangle is RST?
a. equilateral
b. isosceles
c. scalene
d. right
9. Write an equation of the line that contains the median of traingle RST from S to RT .
a. x + 2 y = -10
b. x + 2 y = 6
c. x + 2 y = 8
d. x + 2 y = 10
10. Write the equation of the line that contains the altitude of triangle RST from T to RS .
a. 2x - 3y = -13
b. 3 x -2 y = -2
c. 2 x + 3 y = 29
d. 3 x + 2 y = 26
What is to find an approximate value for a number is called? any words starting with the letter 'r'?
There are 4 quarters in 1 dollar. The total number of dollars is a function of the number of quarters. Does this situation represent a linear or nonlinear function? Explain why.
The situation represents the linear function.
Given that,
There are 4 quarters in one dollar. The total no of dollars represents the function of the no of the quarter.Based on the above information, we can say that
When dollar presents y value and the quarter presents x value
So, the slope should be [tex]\frac{1}{4}[/tex]
The linear function should be [tex]y = \frac{1}{4}x[/tex]
Therefore we can conclude that the situation represents the linear function.
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5.1(x 2)=1.02 how do i solve it
What is the result when -2.5 is divided by 1.8?
n a division problem, the divisor is twenty times the quotient and five times the remainder. If remainder is 16, the number will be:
Option 1 : 3360 Option 2 : 336 Option 3 : 1616 Option 4 : 20516
Factor completely 50a2b5 – 35a4b3 + 5a3b4
Answer
10b2 – 7a2 + ab
a2b3(50b2 – 35a2 + 5ab)
5(10a2b5 – 7a4b3 + a3b4)
5a2b3(10b2 – 7a2 + ab)
...?
Answer:
The factored form of the given expression is [tex]5a^2b^3(10b^2-7a^2+ab)[/tex]
Step-by-step explanation:
We have been given the expression [tex]50a^2b^5-35a^4b^3+5a^3b^4[/tex]
In order to factor it completely we can check for the GCF (greatest common factor) among all the three terms
The GCF is [tex]5a^2b^3[/tex]
On factor out the GCF, we are left with
[tex]5a^2b^3(10b^2-7a^2+ab)[/tex]
Therefore, the factored form of the given expression is [tex]5a^2b^3(10b^2-7a^2+ab)[/tex]
Answer:
given expression is [tex]5a^{2} b^{3} (10b^{2}-7a^{2} +ab)[/tex]
Step-by-step explanation:
Factor and simplify: 4a^2c^2(a^2-b^2+c^2)^2