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Fynjy0 [20]
4 years ago
11

Please help and explain!

Mathematics
1 answer:
Sphinxa [80]4 years ago
5 0
I think I'm not sure sorry if it's wrong

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If you have 12 circles an you shade in3/4 of them, how many circles do you shade?
aleksandrvk [35]
This is a formula if you need one. DON'T forget to cross-multiply

3/4 = x/12


4 0
3 years ago
Write an equation in point-form of the line through point J(4,-4) with slope 4
Ksivusya [100]
Plug into y-y1=m(x-x1)
y+4=4(x-4)


8 0
3 years ago
Name
kozerog [31]

2, 8, 32, 128,

Multiply by 4 for each number

8/2= 4

32/8= 4

128/32= 4

128*4=512

512*4= 2,048

Answer: D 512, 2,048

5 0
4 years ago
Write the equation of a line that passes through the points (−3, 7) and (3, 3)
Dennis_Churaev [7]

Step-by-step explanation:

You can write an equation of a line conveniently by point-slope form. It's in the form of y -a = m(x -b) where (b,a) is the coordinates of a point that's on the line and m is the slope of the line.

Now choose a point (It doesn't really matter which one) and plug that in the equation. I'll choose (-3,7) where a = 7 and b = -3

y -a = m(x -b) \\ y -7 = m(x -(-3)) \\ y -7 = m(x +3)

The next thing we have to do now is finding the slope, m, where it's equal to \frac{y_{2} -y_{1}}{x_{2} -x_{1}}\\. I'll make (-3,7) point 1 and (3,3) point 2.

m = \frac{3 - 7}{3 -(-3)} \\ m = \frac{3 - 7}{3 +3} \\ m = \frac{-4}{6} \\ m = -\frac{2}{3}

Now let's plug that to our equation.

y -7 = m(x +3) \\ y -7 = -\frac{2}{3}(x +3)

Now we have the equation but out of all the choices it seemed that all of them are in slope-intercept form all you have to do now is make our equation rewrite it in slope-intercept form.

y -7 = -\frac{2}{3}(x +3)\\ y -7 = -\frac{2}{3}x -\frac{2}{3}(3) \\ y - 7 = -\frac{2}{3}x -2\\ y = -\frac{2}{3}x +5

<h3>Answer:</h3>

y = -\frac{2}{3}x +5\\ is your equation.

3 0
3 years ago
Y=-x^2+2x+10<br> y=x+2<br><br> Substitution <br> Please show your work<br> Need ASAP
erik [133]

Answer:

The solutions of the system of equations are the points

(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})  

(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})  

Step-by-step explanation:

we have

y=-x^{2} +2x+10 ----> equation A

y=x+2 ----> equation B

Solve the system by substitution

substitute equation B in equation A

x+2=-x^{2} +2x+10

solve for x

-x^{2} +2x+10-x-2=0

-x^{2} +x+8=0

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}

in this problem we have

-x^{2} +x+8=0

so

a=-1\\b=1\\c=8

substitute in the formula

x=\frac{-1\pm\sqrt{1^{2}-4(-1)(8)}} {2(-1)}

x=\frac{-1\pm\sqrt{33}} {-2}

x=\frac{-1+\sqrt{33}} {-2}  -----> x=\frac{1-\sqrt{33}} {2}  

x=\frac{-1-\sqrt{33}} {-2}  -----> x=\frac{1+\sqrt{33}} {2}  

<em>Find the values of y</em>

For x=\frac{1-\sqrt{33}} {2}  

y=x+2

y=\frac{1-\sqrt{33}} {2}+2  ---->y=\frac{5-\sqrt{33}} {2}  

For x=\frac{1+\sqrt{33}} {2}  

y=x+2

y=\frac{1+\sqrt{33}} {2}+2  ---->y=\frac{5+\sqrt{33}} {2}  

therefore

The solutions of the system of equations are the points

(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})  

(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})  

5 0
3 years ago
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