All categories

3 years ago

What is the y-value in the solution to this system of linear equations?

Mathematics

2 answers:

coldgirl [10]3 years ago

8 0

Answer:

A. -4

Step-by-step explanation:

To find the y-value in the solution to this system of linear equation, we will follow the steps below:

4x + 5y = −12 ------------------------------------------------------(1)

-2x + 3y = −16 -----------------------------------------------------(2)

Multiply through equation (1) by 2 and multiply through equation (2) by 4 to make the x coefficient have equal value

8x + 10y = -24 ---------------------------------------------------(3)

-8x + 12y = -64 ---------------------------------------------------(4)

Add equation (1) and equation (2)

22y = -88

Divide both-side of the equation by 22

22y/ 22 = -88/ 22

y =-4

Westkost [7]3 years ago

3 0

Answer:

Step-by-step explanation:

Given the 2 equations

4x + 5y = - 12 → (1)

- 2x + 3y = - 16 → (2)

Eliminate the x- term by multiplying (2) by 2 and adding the result to (1)

- 4x + 6y = - 32 → (3)

Add (1) and (3) term by term

11y = - 44 ( divide both sides by 11 )

y = - 4

You might be interested in

Select all equations that could be used to find the measure of angle C, using the fact that the sum of all the angles in a trian

Semmy [17]

C = 180 - A - B

And

C = 180 - (A+B)

Hope I can help you :)
Brainliest answer :)) ?

4 0

3 years ago

Evaluate the line integral, where c is the given curve. c y3 ds,

postnew [5]

Parameterizing $\mathcal C$ by

$\mathbf r(t)=(t^3,t)$

with $0\le t\le2$ , we have

$\mathrm ds=\|\mathbf r'(t)\|\,\mathrm dt=\sqrt{(3t^2)^2+1^2}\,\mathrm dt=\sqrt{9t^4+1}\,\mathrm dt$

So

$\displaystyle\int_{\mathcal C}y^3\,\mathrm ds=\int_{t=0}^{t=2}t^3\sqrt{9t^4+1}\,\mathrm dt$
$=\displaystyle\frac1{36}\int_0^236t^3\sqrt{9t^4+1}\,\mathrm dt$
$=\displaystyle\frac1{36}\int_{u=1}^{u=145}\sqrt u\,\mathrm du$
$=\dfrac{145^{3/2}-1}{54}$

5 0

3 years ago

Use a translation rule to describe the translation of ABC that is 6 units to the right and 10 units down.

vesna_86 [32]

Answer:

(x, y ) → (x + 6, y - 10 )

Step-by-step explanation:

6 units right is + 6 in the x- direction

10 units down is - 10 in the y- direction

the translation rule is

(x, y ) → (x + 6, y - 10 )

6 0

2 years ago

What is the length of leg sof the triangle below?<br> 90°

miskamm [114]

Not sure I’m sorry, have you tried google

5 0

3 years ago

A college requires applicants to have an ACT score in the top 12% of all test scores. The ACT scores are normally distributed, w

DochEvi [55]

Answer:

a) The lowest test score that a student could get and still meet the colleges requirement is 27.0225.

b) 156 would be expected to have a test score that would meet the colleges requirement

c) The lowest score that would meet the colleges requirement would be decreased to 26.388.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 21.5, \sigma = 4.7$

a. Find the lowest test score that a student could get and still meet the colleges requirement.

This is the value of X when Z has a pvalue of 1 - 0.12 = 0.88. So it is X when Z = 1.175.

$Z = \frac{X - \mu}{\sigma}$

$1.175 = \frac{X - 21.5}{4.7}$

$X - 21.5 = 1.175*4.7$

$X = 27.0225$

The lowest test score that a student could get and still meet the colleges requirement is 27.0225.

b. If 1300 students are randomly selected, how many would be expected to have a test score that would meet the colleges requirement?

Top 12%, so 12% of them.

0.12*1300 = 156

156 would be expected to have a test score that would meet the colleges requirement

c. How does the answer to part (a) change if the college decided to accept the top 15% of all test scores?

It would decrease to the value of X when Z has a pvalue of 1-0.15 = 0.85. So X when Z = 1.04.

$Z = \frac{X - \mu}{\sigma}$

$1.04 = \frac{X - 21.5}{4.7}$

$X - 21.5 = 1.04*4.7$

$X = 26.388$

The lowest score that would meet the colleges requirement would be decreased to 26.388.

6 0

3 years ago