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3 years ago

25 POINTS HELPPPP!!! solve question 1, 2, 7, 8 on the picture please

Mathematics

2 answers:

Snowcat [4.5K]3 years ago

7 0

1. x=12

2. y= -5

7. y= -4

8. x= - 3/10

dem82 [27]3 years ago

6 0

Answer:

1. B) x = 12

2.C) -5

3. A) 4

7. C) -4

8. A) -4x -6 = 16x

9. B) 29x = 0

Step-by-step explanation:

okay, number 7 you have to distribute -3 to 6 to get -18. So you get 4 -(3*2y) - 18 = 10. add 18 to both sides to get 4 -(3*2y) = 28. subtract the 4. -(3*2y) = 24. Rever the negative. (3*2y) = -24. Divid by 3; 2y = -8. And divide by 2; y = -4.

Same with number 8 but keeping the 16x.

4x - 2 (4x + 3) = 16x

4x - 8x + - 6 = 16x

-4x + -6 = 16x

4(3 - 4x) = 12

3 - 4x = 3

-4x = 0

x = 0

Now you know x = 0 so anything multipied

with the x is 0. The answer is B.

I hope this helped :)

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Read 2 more answers

Write an equation for an ellipse centered at the origin, which has foci at (\pm\sqrt{12},0)(± 12 ,0)left parenthesis, plus min

lora16 [44]

Answer:

$\mathbf{\dfrac{x^2}{49^2} +\dfrac{y^2}{37^2} =1}$

Step-by-step explanation:

Given that :

the foci of the ellipse is (±√12,0) and C0-vertices are (0,±√37)

The foci are (-C,0) and (C ,0)

the focus has x-coordinates so the focus is lie on x- axis.

The major axis also lie on x-axis

The minor axis lies on y-axis so C0-vertices are (0,±√37)

The given focus C = ae = √12

Given co-vertices ( minor axis) (0,±b) = (0,±√37)

b= √37

We can therefore express the relation between the focus and semi major axes and semi minor axes as:

$\mathbf{c^2 = a^2 - b^2 } \\ \\ \mathbf{a^2 = c^2 + b^2 } \\ \\ \mathbf{c^2 = ( \sqrt12)^2 - (\sqrt 37)^2 } \\ \\ \mathbf{c^2 = 49 } \\ \\ \mathbf{c = \sqrt{49 }}$

The equation of ellipse formula is:

$\dfrac{x^2}{a^2} +\dfrac{y^2}{b^2} =1$

and we know that $\mathbf{a=\sqrt{49} \ \ and \ \ b=\sqrt{37}}$

Thus ; the equation of the ellipse at the origin is

$\mathbf{\dfrac{x^2}{49^2} +\dfrac{y^2}{37^2} =1}$

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4 years ago