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3 years ago

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The vertices of a hyperbola are located at (0, –4) and (0, 12). The foci of the same hyperbola are located at (0,–6) and (0, 14)

. What is the equation of the hyperbola?

2 answers:

Drupady [299]3 years ago

7 0

Answer:

correct asnwer is C

Step-by-step explanation:

Norma-Jean [14]3 years ago

5 0

Answer:

The Answer is C on edge

Step-by-step explanation:

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3 years ago

Solving Simultaneously equations<br> 3. 2x + 3y = 1<br> -x + 2y = - 4

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2x5y7

Step-by-step explanation:

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2 years ago

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A^2 + b^2 + c^2 = 2(a − b − c) − 3. (1) Calculate the value of 2a − 3b + 4c.

Verdich [7]

Answer:

$2a - 3b + 4c = 1$

Step-by-step explanation:

Given

$a^2 + b^2 + c^2 = 2(a - b - c) - 3$

Required

Determine $2a - 3b + 4c$

$a^2 + b^2 + c^2 = 2(a - b - c) - 3$

Open bracket

$a^2 + b^2 + c^2 = 2a - 2b - 2c - 3$

Equate the equation to 0

$a^2 + b^2 + c^2 - 2a + 2b + 2c + 3 = 0$

Express 3 as 1 + 1 + 1

$a^2 + b^2 + c^2 - 2a + 2b + 2c + 1 + 1 + 1 = 0$

Collect like terms

$a^2 - 2a + 1 + b^2 + 2b + 1 + c^2 + 2c + 1 = 0$

Group each terms

$(a^2 - 2a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

Factorize (starting with the first bracket)

$(a^2 - a -a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$(a(a - 1) -1(a - 1)) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1) (a - 1)) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b^2 + b+b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b(b + 1)+1(b + 1)) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)(b + 1)) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c^2 + c+c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c(c + 1)+1(c + 1)) = 0$

$((a - 1)^2) + ((b + 1)^2) + ((c + 1)(c + 1)) = 0$

$((a - 1)^2) + ((b + 1)^2) + ((c + 1)^2) = 0$

Express 0 as 0 + 0 + 0

$(a - 1)^2 + (b + 1)^2 + (c + 1)^2 = 0 + 0+ 0$

By comparison

$(a - 1)^2 = 0$

$(b + 1)^2 = 0$

$(c + 1)^2 = 0$

Solving for $(a - 1)^2 = 0$

Take square root of both sides

$a - 1 = 0$

Add 1 to both sides

$a - 1 + 1 = 0 + 1$

$a = 1$

Solving for $(b + 1)^2 = 0$

Take square root of both sides

$b + 1 = 0$

Subtract 1 from both sides

$b + 1 - 1 = 0 - 1$

$b = -1$

Solving for $(c + 1)^2 = 0$

Take square root of both sides

$c + 1 = 0$

Subtract 1 from both sides

$c + 1 - 1 = 0 - 1$

$c = -1$

Substitute the values of a, b and c in $2a - 3b + 4c$

$2a - 3b + 4c = 2(1) - 3(-1) + 4(-1)$

$2a - 3b + 4c = 2 +3 -4$

$2a - 3b + 4c = 1$

7 0

3 years ago

ASAP please help!!!

Serggg [28]

Y+4<u><</u>9
subtract 4
y<u><</u>5
y is smaller than and equal to 5
so you shade from 5 to the negative end to infinity (to the left) and shade the 5 to show that it is included (attachment says A)

6 less than (-6) 2 times a number (2 time x) is greater than (>) 8 (8)

-6+2x>8
add 6
2x>14
divide 2
x>7
so
x is bigger than 7
shade from 7 to the positive end to infinity (to the right) and don't shade 7 but put a circle around it to show that it is not included (attachment says B )

I have included pictures of the number lines

4 0

3 years ago

Which equation is part of solving the system by substitution? 4(y + 11)2 – 3y2 = 8 4(11 – y)2 – 3y2 = 8 4(y – 11)2 – 3y2 = 8 4(–

swat32

Answer:

$4(11-y)^{2} -3y^{2}=8$

Step-by-step explanation:

we have

$x+y=11$ ----> equation A

$4x^{2} -3y^{2}=8$ ----> equation B

Solve the system by substitution

step 1

isolate the variable x in the equation A

$x=11-y$ ----> equation A1

step 2

Substitute the equation A1 in the equation B

$4(11-y)^{2} -3y^{2}=8$

Solve for y

3 0

3 years ago

Read 2 more answers