The sum of two consecutive integers is 59. Write an equation that models this situation and find the values of the two integers.
A) n + 2n = 59; n = 19; 2n = 38
B) n + n + 1 = 59; n = 29; n + 1 = 28
C) n + n + 1 = 59; n = 30; n + 1 = 31
D) n + n + 1 = 59; n = 29; n + 1 = 30
Please explain your answer! : )
Liam earns $7.50 an hour. His benefits package is equal to 25 percent of his hourly wages. When you include the value of his benefits, how much does Liam earn per hour?
Micheala mixed peanuts and almonds. · Micheala can buy 3 pounds of peanuts for a total cost of $8.70. · The cost per pound for almonds is 80% more than the cost per pound for peanuts. · Micheala bought enough peanuts that, when he mixed them with the almonds, the mixture had a value of $3.94 per pound. What is the approximate percent by weight of almonds?
Find the polynomial.
{-1/3, 4} is the solution set of?
A. 3x^2 - 11x + 4 = 0
B. 3x^2 - 11x - 4 = 0
C. 1/3x^2 - 11x - 4 = 0
D.-1/x^2 - 11x - 4 = 0
Answer:
option B
Step-by-step explanation:
A. [tex]3x^2 - 11x + 4 = 0[/tex]
3*4 = 12
We find out two factors whose sum is -11 and product is 12
1 times 12 = 12
To get sum -11 , then one factor should be negative. So, factoring is not possible.
B . [tex]3x^2 - 11x - 4 = 0[/tex]
3*(-4) = -12
We find out two factors whose sum is -11 and product is -12
1 times (-12) = -12
1 + (-12) = -11
So two factors are 1 and -12
Split the middle term -11x using factors 1 and -12
So equation becomes
[tex]3x^2 + 1x - 12x - 4 = 0[/tex]
Now group first two terms and last two terms
[tex](3x^2 + 1x)+ (- 12x - 4) = 0[/tex]
[tex]x(3x+ 1)-4(3x +1)=0[/tex]
(3x+1)(x-4)=0
Now we set each parenthesis =0 and solve for x
3x+1 =0 , subtract 1 on both sides
3x = -1 ( divide both sides by 3)
x= -1/3
Now we set x-4=0
add 4 on both sides
so x=4
Option B is correct
Solve the quadratic equation by taking square roots. 3x2 − 4 = 28
You ride your bicycle at a rate of 10 mi/h. the distance d (in miles) that you ride is given by the function d=10t where t is the time in hours.
The function d = 10t signifies that you cover 10 miles of distance per hour.
What is the general equation of a Straight line?The general equation of a straight line is -
[y] = [m]x + [c]
where -
[m] is slope of line which tells the unit rate of change of [y] with respect to [x].
[c] is the y - intercept i.e. the point where the graph cuts the [y] axis.
The equation of a straight line can be also written as -
Ax + By + C = 0
By = - Ax - C
y = (- A/B)x - (C/A)
We have the following problem -
You ride your bicycle at a rate of 10 mi/h. the distance d (in miles) that you ride is given by the function d=10t.
We have the distance [d] as the function of time as -
d = 10t
The given equation id d = 10t. The slope of the line is 10. This means that you cover 10 miles of distance in 1 hour. The graph representing the function is plotted.
Therefore, the function signifies that you cover 10 miles of distance per hour.
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An isosceles triangle’s altitude will bisect its base. Which expression could be used to find the area of the isosceles triangle above?
The expression could be used to find the area of the isosceles triangle above is [tex]\sqrt{40} \cdot \sqrt{40} /2[/tex]
The length of the base is the distance between the points 4+2i and 10+4i, so
Base= |10+ 4i (4+2i)| = |10+4i-4-2i|= |6 + 2i| = [tex]\sqrt{6^2 +2^2}[/tex] = [tex]\sqrt{36+4}[/tex]= √40
The middle point of the base is placed at point
4+2i+ 10 + 4i/2 = 6i +14/2 = 7+ 3i
The length of the height is the distance between the points 5+9i and 7+3i
Height = 5 +91 (7+3i)| =|5+ 9i −7 - 3i| = |−2+6i| = [tex]\sqrt{(-2)^2 + 6^2} = \sqrt {4+36}[/tex] = √40
So, the area of the triangle is
[tex]A= 1/2 \cdot Base \cdot Height= \sqrt{40} \cdot \sqrt{40} /2[/tex]
Therefore, The expression could be used to find the area of the isosceles triangle above is [tex]\sqrt{40} \cdot \sqrt{40} /2[/tex]
The probable question may be:
An isosceles triangle’s altitude will bisect its base. Which expression could be used to find the area of the isosceles triangle above?
Points on the graph of the triangle are (5+9i), (10+4i), and (4+2i).
A. \sqrt{40} \cdot \sqrt{40} /2
B. \sqrt{40} \cdot \sqrt{68} /2
C. \sqrt{232} \cdot \sqrt{288} /2
D. \sqrt{232} \cdot \sqrt{164} /2
Final answer:
To find the area of an isosceles triangle when given the length of the base and the altitude, you can use the formula A = ½ × base × height. The altitude of an isosceles triangle will bisect its base, so you can divide the base in half to find the length of the base above the altitude.
Explanation:
An isosceles triangle's altitude will bisect its base. To find the area of the isosceles triangle, we can use the formula A = ½ × base × height. Since the altitude bisects the base, we can divide the base in half to find the length of the base above the altitude. Let's say the length of the base is 2x and the length of the altitude is h. So, the expression to find the area of the isosceles triangle is A = ½ × (2x) × h = xh.
Slope is -2 and (5, 3) is on the line. What is the equation in slope intercept form?
At what point on the paraboloid y = x2 + z2 is the tangent plane parallel to the plane 3x + 2y + 7z = 2? (if an answer does not exist, enter dne.) (x, y, z) =
Final answer:
The point on the paraboloid y = x^2 + z^2 where the tangent plane is parallel to the plane 3x + 2y + 7z = 2 is found by setting the normal vectors proportional. Solving the equations, the point is (3/4, 29/8, 7/4).
Explanation:
To find the point on the paraboloid y = x2 + z2 where the tangent plane is parallel to the plane 3x + 2y + 7z = 2, we first need to determine the normal vector of the given plane. The normal vector of the plane is defined by its coefficients, which are (3, 2, 7). For the paraboloid, we can find the normal vector at any point by taking the gradient of the function y.
The gradient of y with respect to x and z is (2x, 1, 2z). A tangent plane to the paraboloid at point (x, y, z) will have this gradient as its normal vector. To find the point where this tangent plane is parallel to the given plane, we set the gradients equal to each other up to a constant factor because parallel planes have proportional normal vectors.
Therefore, we solve the equations:
2x = 3k
1 = 2k
2z = 7k
From the second equation, k = 1/2. Substituting k into the other equations, we find x = 3/4 and z = 7/4. Now we can substitute x and z into the equation of the paraboloid to find y:
y = (3/4)2 + (7/4)2 = 9/16 + 49/16 = 58/16 = 29/8.
The point on the paraboloid where the tangent plane is parallel to the given plane is (3/4, 29/8, 7/4).
Mr. lance keeps a jar of candy on his desk for his students to pick from when they do well on a test. The jar contain 5 snickers, 2 butterfingers, 4 almond joys and 3 milky ways. If two students get to pick candy from the jar, what is the probability that the first student picks a snickers and then a second student also picks a snickers?
The probability that the first student picks a Snickers and then the second student also picks a Snickers is 10/91.
Explanation:The question is asking for the probability that two students will each pick a Snickers from a jar containing various types of candy. We start with a total of 14 candies: 5 Snickers, 2 Butterfingers, 4 Almond Joys, and 3 Milky Ways. When the first student picks a Snickers, there are 5 possible Snickers they can pick out of the 14 total candies.
The probability for the first student is therefore rac{5}{14}. Assuming they pick a Snickers, there are now 13 candies left in total, with 4 being Snickers. So, the second student now has a probability of rac{4}{13} to pick a Snickers.
To find the overall probability of both events happening, we multiply the probabilities of each individual event occurring. The calculation is:
rac{5}{14} \times rac{4}{13} = rac{20}{182} = rac{10}{91}
The probability that the first student picks a Snickers and then the second student also picks a Snickers is rac{10}{91}.
Find the area of the parallelogram when base = 10yd and height = 5yd. A. 15yd B. 15yd2 C. 50yd D. 50yd2
area = l x w
10 * 5 = 50
answer is D. 50 yd^2
1. Which domain restrictions apply to the rational expression?
x^2+5x+6
--------------
x^2-9
Select each correct answer.
x≠2
x≠1
x≠−3
x≠3
x≠−2
x≠−1
(4x)/(x-3)+(2)/(x^(2)-9)= (1)/(x+3)
A. 19/3
B. 13/3
C. -5
D. -17/3
9. Which statement best reflects the solution(s) of the equation?
1/x + 1/x-3 = x-2 / x-3
A. There is only one solution: x = 1.
The solution x = 0 is an extraneous solution.
B. There is only one solution: x = 1.
The solution x = 3 is an extraneous solution.
C. There is only one solution: x = 3.
The solution x = 1 is an extraneous solution.
D. There are two solutions: x = 1 and x = 3.
10. (√3x+1)−x+3=0
A. There is only one solution: x = 1.
The solution x = 8 is an extraneous solution.
B. There is only one solution: x = 8.
The solution x = 1 is an extraneous solution.
C. There is only one solution: x = 8.
The solution x = 0 is an extraneous solution.
D. There are two solutions: x = 1 and x = 8.
Answer with explanation:
1. The given rational expression is
[tex]\rightarrow\frac{x^2+5 x+6}{x^2-9}\\\\\rightarrow\frac{x^2+5 x+6}{(x-3)(x+3)}[/tex]
The function is not defined ,when
→(x-3)(x+3)=0
→x-3≠0 ∧ x+3≠0
→x≠3, ∧ x ≠ -3
⇒Option C and D
→x≠−3
→x≠3
2.
[tex]\rightarrow\frac{4 x}{x-3}+\frac{2}{x^2-9}=\frac{1}{x+3}\\\\\rightarrow\frac{4 x}{x-3}+\frac{2}{(x-3)(x+3)}=\frac{1}{x+3}\\\\\rightarrow\frac{4 x(x+3)+2}{(x-3)(x+3)}=\frac{1}{x+3}\\\\\rightarrow4 x(x+3)+2=\frac{(x-3)(x+3)}{x+3}\\\\\rightarrow4x^2+12 x+2=x-3\\\\\rightarrow4x^2+11x+5=0\\\\ \text{Using Discriminant method for a quadratic equation}\\\\ax^2+bx +c=0\\\\x=\frac{-b\pm\sqrt{D}}{2 a}\\\\D=b^2-4 ac\\\\x=\frac{-11\pm\sqrt{121-80}}{2 \times 4}\\\\x=\frac{-11\pm\sqrt{41}}{8}[/tex]
None of the option
3.
[tex]\rightarrow \frac{1}{x}+\frac{1}{x-3}=\frac{x-2}{x-3}\\\\\rightarrow\frac{x-3+x}{x(x-3)}=\frac{x-2}{x-3}\\\\\rightarrow2x-3=\frac{x(x-3)(x-2)}{x-3}\\\\\rightarrow 2 x-3=x^2-2 x\\\\\rightarrow x^2-4x+3=0\\\\\rightarrow (x-1)(x-3)=0\\\\x=1,3[/tex]
For, x=3 , the equation is not defined.
So, there is single solution which is , x=1.
Option B:→ There is only one solution: x = 1.
The solution x = 3 is an extraneous solution.
4.
[tex]\rightarrow \sqrt{3}x+1-x+3=0\\\\\rightarrow \sqrt{3}x -x=-4\\\\\rightarrow x(\sqrt{3}-1)=-4\\\\\rightarrow x=\frac{-4}{\sqrt{3}-1}\\\\\rightarrow x=\frac{-4\times(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}\\\\x=\frac{-4\times(\sqrt{3}+1)}{2}\\\\x=-2(\sqrt{3}+1)[/tex]
None of the option
1) Options c) and d) are correct.
2) None of the options are correct.
3) Option B) is correct.
4) None of the options are correct.
Step-by-step explanation:
1) Given : [tex]4x^2+12x+2=x-3[/tex]
Expression -- [tex]\dfrac{x^2+5x+6}{x^2-9}[/tex]
Solution :
[tex]\dfrac{x^2+5x+6}{x^2-9}=\dfrac{x^2+3x+2x+6}{(x+3)(x-3)}[/tex]
[tex]\dfrac{x^2+5x+6}{x^2-9}=\dfrac{(x+3)(x+2)}{(x+3)(x-3)}[/tex]
Therefore, [tex]\rm x\neq 3 \; and\;x\neq -3[/tex] ,option c) and d) is correct.
2) Given :
Expression - [tex]\dfrac{4x}{x-3}+\dfrac{2}{x^2-9}=\dfrac{1}{x+3}[/tex]
Solution :
[tex]\dfrac{4x}{(x-3)}+\dfrac{2}{(x+3)(x-3)}=\dfrac{1}{(x+3)}[/tex]
[tex]\dfrac{4x(x+3)+2}{(x-3)(x+3)}=\dfrac{1}{(x+3)}[/tex]
[tex]4x^2+12x+2=x-3[/tex]
[tex]4x^2+11x+5=0[/tex]
[tex]x=\dfrac{-11\pm\sqrt{121-80} }{8}[/tex]
[tex]x = \dfrac{-11\pm\sqrt{41} }{8}[/tex]
None of the options are correct.
3) Given :
Expression - [tex]\dfrac{1}{x}+\dfrac{1}{x-3}=\dfrac{x-2}{x-3}[/tex] ----- (1)
Solution :
[tex]\dfrac{x-3+x}{(x)(x-3)}=\dfrac{x-2}{x-3}[/tex]
[tex]2x-3=x(x-2)[/tex]
[tex]x^2-4x+3=0[/tex]
[tex]x^2-3x-x+3=0[/tex]
[tex](x-3)(x-1)=0[/tex]
At x = 3 equation (1) is not define. Therefore, the correct answer is option
B) There is only one solution: x = 1. The solution x = 3 is an extraneous solution.
4) Given :
Exprression - [tex](\sqrt{3}x +1)-x+3=0[/tex]
Solution :
[tex]x(\sqrt{3}-1 )=-4[/tex]
[tex]x=\dfrac{-4}{\sqrt{3}-1 }[/tex]
None of the options are correct.
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a-b+2c=7
b+c=1
a-2b-3c=0
translate the word phrase nto a variable expression the quotient of a number and 3 is increased by 1
HELPPPPPPPPPPPPPPPPPPPPP
Consider the graph of the linear function h(x) = –x + 5. Which could you change to move the graph down 3 units?
-3(y+4)=21
solve for y
show all work please , thank u
If the perimeter of the rectangle is 28 cm, find the value(s) for x.
A donkey weighs 570 pounds and an elephant weighs 5 tons. How much more does the elephant weigh?
The correct answer is:
5 tons = 10,000 lb
10,000 lb − 570 lb = 9,430 lb = 4 tons and 1,430 lb
This answer is if you want it in tons and pounds.
need help from a math wiz
The sum of the interior angle measures of a convex polygon is 1440°. how many sides does it have?
Which statement about the equation is true? 3y – 1 = 1/3 – 4y
The equation has one solution.
The equation has no solution.
The equation has a few solutions.
The equation has many solutions.
Se the change of variables s=xy, t=xy^2 to compute \iint_r xy^2\,da, where r is the region bounded by xy=3,\ xy=7,\ xy^2=3,\ xy^2=7.
The value of [tex]\int\limits^._R {xy^2} \, dA[/tex] is 16.
What is integration?The calculation of an integral is called integration. In math, many useful quantities like areas, volumes, displacement, and so on can be found using integrals. When we discuss integrals, we typically refer to definite integrals. Antiderivatives make use of the indefinite integrals. Apart from differentiation, one of the two major calculus topics in mathematics is integration.
Given xy =3, xy = 7,
xy² = 3, xy² = 7
s = xy and t = xy²
dividing t by s we get
t/s = xy²/xy
y = t/s
and x = s/y = s²/t
now differentiate x and y partially with respect to s and t
∂x/∂s = 2s/t
∂x/∂t = -s²/t²
∂y/∂s = -t/s²
∂y/∂t = 1/s
The Jacobian is
∂(x, y)/∂(s, t) = [tex]\left|\begin{array}{ccc}dx/ds&dx/dt\\dy/ds&dy/dt\end{array}\right|[/tex]
∂(x, y)/∂(s, t) =[tex]\left|\begin{array}{ccc}2s/t&-s^{2}/t^{2} \\-t/s^{2} &1/s\end{array}\right|[/tex]
∂(x, y)/∂(s, t) =2/t - 1/t
∂(x, y)/∂(s, t) = 1/t
so for [tex]\int\limits^._R {xy^2} \, dA[/tex] = [tex]\int\limits^a_b \int\limits^a_b {t}\frac{d(x, y)}{d(s, t) } \, dsdt[/tex]
for (s, t) (3 ≤ s ≤7, 3 ≤ t ≤ 7)
= [tex]\int\limits^7_3 \int\limits^7_3 {t}*1/t } \, dsdt[/tex]
= [tex]\int\limits^7_3ds \int\limits^7_3 dt[/tex]
=[s]₃⁷ [t]₃⁷
= (7 - 3)(7 - 3)
= 16
Hence the value is 16.
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are these points correct??
What is the solution to 89+k=-42
Which one of the following is a perfect square A: 100, B:13, C:127, or D:102
PLEASE HELP ME!!!!! WORTH 10 POINTS!!!!
Part 1: Use the graph to count the slope of the line that passes through the points (2, 1) and (2, 0).
Part 2: In two or more complete sentences, explain why it is not possible to write the equation of the given line in the traditional version of the point-slope form of a line.
Answer:
Notice that the given line is vertical, which means its slope is undetermined.
We can demonstrate that using the slope definition:
[tex]m=\frac{y_{2} -y_{1} }{x_{2}-x_{1} }[/tex]
[tex]m=\frac{1-0}{2-2}=\frac{1}{0}=IND[/tex]
Which is undetermined because null denominators cannot be determined.
Now, the Point-Slope form of a line refers to expressin the equation with one point and its slope. So, if we cannot determine the slope, we cannot determine the point-slope form neither. In other words, we need to have a determined slope to express it in this form.
If the interior angle of a polygon is 170 how many sides does it have
if interior angle is 170, subtract 170 from 180 to find exterior angle:
180-170 = 10 = exterior angle
the sum of exterior angles is 360 so divide by exterior angle to find sides:
360 /10 = 36 sides
Line g bisects line segment BC at Point H. BH = 4x and HC = 24. Find BC.
Answer:
The length of BC is 48 units.
Step-by-step explanation:
Given information: Line g bisects line segment BC at Point H. BH = 4x and HC = 24.
It is given that line g bisects line segment BC at Point H. It means point H divides the line BC in two equal parts BH and HC.
[tex]BH=HC[/tex]
[tex]4x=24[/tex]
Divide both sides by 4.
[tex]x=6[/tex]
The value of x is 6.
[tex]BH=4x=6\times 4=24[/tex]
The length of BC is
[tex]BC=BH+HC[/tex] (Segment addition property)
[tex]BC=24+24=48[/tex]
Therefore the length of BC is 48 units.
Is it possible to draw a graph of this system? If yes, draw it.
[tex] \left \{ {{3x-1=0} \atop {4x+2y=0}} \right. [/tex]
Thanks in adavance!