How far apart are the foci of an ellipse with a major axis of 20 feet and a minor axis of 14 feet?

1 answer:

3 0

The solution for this question is simple. First we need to divide both numbers by 2 which will give us 10 and 7. This is the equation x^2/10^2 + y^2/7^2 = 1sqrt(100-49) = 7.14 (the answer do not have a perfect square root)
Foci are 14.28 units apart

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What is the value of x in -4(x+3)=8<br><br> please help me quickly !!

xz_007 [3.2K]

Step 1: Simplify both sides of the equation.

−4(x+3)=8

(−4)(x)+(−4)(3)=8(Distribute)

−4x+−12=8

−4x−12=8

Step 2: Add 12 to both sides.

−4x−12+12=8+12

−4x=20

Step 3: Divide both sides by -4.

−4x

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−4

________

−4

x=−5

Answer:

x=−5

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2 years ago

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morpeh [17]

Answer:

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2 years ago

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Find the distance of the line segment joining the two points (-4 /2 - /12) and (/32, 2/3)

astraxan [27]

Answer: $4\sqrt{3}$ .

Step-by-step explanation:

Distance formula : Distance between points (a,b) and (c,d) is given by :-

$D=\sqrt{(d-b)^2+(b-a)^2}$

Distance between points $(-4\sqrt{2},\sqrt{12}) \text{ and }(-\sqrt{32}, 2\sqrt{3})$ .

$D=\sqrt{(2\sqrt{3}-(-\sqrt{12}))^2+(-\sqrt{32}-(-4\sqrt{2}))}\\\\=\sqrt{(2\sqrt{3}+\sqrt{2\times2\times3})^2+(-\sqrt{4\times4\times2}+4\sqrt{2})^2}\\\\=\sqrt{(2\sqrt{3}-\sqrt{2^2\times3})^2+(-\sqrt{4^2\times2}+4\sqrt{2})^2}\\\\=\sqrt{(2\sqrt{3}+2\sqrt{3})^2+(-4\sqrt{2}+4\sqrt{2})^2}\\\\=\sqrt{(4\sqrt{3})^2+0}\\\\=4\sqrt{3}\text{ units}$