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3 years ago

Brainers, please help me please

Mathematics

1 answer:

ra1l [238]3 years ago

4 0

Answer:

W = 12

L = 16

Step-by-step explanation:

Givens

L = L

W = 1/2 L + 4

Perimeter = 56

Formula

2L + 2W = Perimeter

Solution

Sustitute

2L + 2(1/2L + 4) = 56

2L + L + 8 = 56

3L + 8 = 56

3L + 8 - 8 = 56 - 8

3L = 48

3L/3 = 48/3

L = 16

W = 1/2 L + 4

W = 8 + 4

W = 12

Check

2L + 2W = 56

2*16 + 2*12 = 56

32 + 24 = 56

56 = 56 and it checks.

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3 years ago

Write the equation for the hyperbola with foci (–12, 6), (6, 6) and vertices (–10, 6), (4, 6).

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Answer:

$\frac{(x--3)^2}{49} -\frac{(y-6)^2}{32}=1$

Step-by-step explanation:

The standard equation of a horizontal hyperbola with center (h,k) is

$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$

The given hyperbola has vertices at (–10, 6) and (4, 6).

The length of its major axis is $2a=|4--10|$ .

$\implies 2a=|14|$

$\implies 2a=14$

$\implies a=7$

The center is the midpoint of the vertices (–10, 6) and (4, 6).

The center is $(\frac{-10+4}{2},\frac{6+6}{2}=(-3,6)$

We need to use the relation $a^2+b^2=c^2$ to find $b^2$ .

The c-value is the distance from the center (-3,6) to one of the foci (6,6)

$c=|6--3|=9$

$\implies 7^2+b^2=9^2$

$\implies b^2=9^2-7^2$

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We substitute these values into the standard equation of the hyperbola to obtain:

$\frac{(x--3)^2}{7^2} - \frac{(y-6)^2}{32}=1$

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