25% of students in 8th grade play an instrument. How many students play an instrument if there are 200 students in 8th grade.

Question 7 Multiple Choice Worth 1 points)

expeople1 [14]

Answer:

The answer to your question is (3, 0)

Step-by-step explanation:

Data

4x + 2y = 12 Equation l

x - y = 3 Equation ll

Process

To solve this problem use the elimination method

1.- Multiply equation ll by 2

4x + 2y = 12

2x - 2y = 6

6x 0 = 18

2.- Solve for x

6x = 18

x = 18/6

x = 3

3.- Substitute x in equation ll to find y

3 - y = 3

-Solve for y

-y = 3 - 3

y = 0

4.- Write the solution

(3, 0)

8 0

3 years ago

Read 2 more answers

If you answer correctly I will give you brainlest GOES TO THE FIRST PERSON WHO ANSWERS CORRECTLY!

Bogdan [553]

Answer:

A) slope is -12

B)The rate of change is constant because for every 5 seconds the change is the same. For every 5 seconds, the height decreases by 60ft. The equation is y=12x+1210

Step-by-step explanation:

6 0

3 years ago

The probability of an adverse reaction to the flu shot is 0.15. Let r be the number of people stuffering a reaction from a flu s

shtirl [24]

Stuffering?
what are you asking???

5 0

3 years ago

I have 2 possible answer, but I don't know which one. Help please.

stira [4]

Answer:

X=120

Step-by-step explanation:

It is an equalateral so all the angles in the middle are the same all being 60

The angle of a straight line is 180

So just minus 60 from 180 to get 120

180-60=120

5 0

3 years ago

Read 2 more answers

Write the equation -4x^2+9y^2+32x+36y-64=0 in standard form. Please show me each step of the process!

IgorC [24]

Hey there, hope I can help!

$-4x^2+9y^2+32x+36y-64=0$

$\mathrm{Add\:}64\mathrm{\:to\:both\:sides} \ \textgreater \ 9y^2+32x+36y-4x^2=64$

$\mathrm{Factor\:out\:coefficient\:of\:square\:terms} \ \textgreater \ -4\left(x^2-8x\right)+9\left(y^2+4y\right)=64$

$\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}4$
$-\left(x^2-8x\right)+\frac{9}{4}\left(y^2+4y\right)=16$

$\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}9$
$-\frac{1}{9}\left(x^2-8x\right)+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}$

$\mathrm{Convert}\:x\:\mathrm{to\:square\:form}$
$-\frac{1}{9}\left(x^2-8x+16\right)+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)$

$\mathrm{Convert\:to\:square\:form}$
$-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)$

$\mathrm{Convert}\:y\:\mathrm{to\:square\:form}$
$-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y^2+4y+4\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right)$

$\mathrm{Convert\:to\:square\:form}$
$-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y+2\right)^2=\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right)$

$\mathrm{Refine\:}\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right) \ \textgreater \ -\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y+2\right)^2=1$

$Refine\;once\;more\;-\frac{\left(x-4\right)^2}{9}+\frac{\left(y+2\right)^2}{4}=1$

For me I used
$\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}= 1$
$As\;\mathrm{it\;\:is\:the\:standard\:equation\:for\:an\:up-down\:facing\:hyperbola}$

I know yours is an equation which is why I did not go any further because this is the standard form you are looking for. I would rewrite mine to get my hyperbola standard form. However the one I have provided is the form you need where mine would be.
$\frac{\left(y-\left(-2\right)\right)^2}{2^2}-\frac{\left(x-4\right)^2}{3^2}=1$

Hope this helps!

4 0

4 years ago