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vekshin1
3 years ago
7

Using the given information, give the vertex form equation of each parabola.

Mathematics
1 answer:
WITCHER [35]3 years ago
5 0

Answer:

Step-by-step explanation:

If you plot these points on a coordinate plane, you will see that they both lie on the same horiztonal line, y = -10, with the focus 1/4 to the right of the vertex.  This information tells you a ton of stuff.  First, it tells you that, since a parabola "hugs" the focus, this parabola opens to the right and is of the form x = y-squared.  This equation is

4p(x-h)=(y-k)^2

The information also tells us that p (the distance between the vertex and the focus) is .25.  h is the "x" coordinate of the vertex and k is the "y" coordinate of the vertex, so h = -1, and k = -10.  Filling in our formula:

4(.25)(x+1)=(y+10)^2

Simplify the left to

1(x+1)=(y+10)^2 and solve for x:

x=(y+10)^2-1

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2 years ago
Based on Pythagorean identities, which equation is true? A. Sin^2 theta -1= cos^2 theta B. Sec^2 theta-tan^2 theta= -1 C. -cos^2
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Answer:

D

Step-by-step explanation:

our basic Pythagorean identity is cos²(x) + sin²(x) = 1

we can derive the 2 other using the listed above.

1. (cos²(x) + sin²(x))/cos²(x) = 1/cos²(x)

1 + tan²(x) = sec²(x)

2.(cos²(x) + sin²(x))/sin²(x) = 1/sin²(x)

cot²(x) + 1 = csc²(x)

A. sin^2 theta -1= cos^2 theta

this is false

cos²(x) + sin²(x) = 1

isolating cos²(x)

cos²(x) = 1-sin²(x), not equal to sin²(x)-1

B. Sec^2 theta-tan^2 theta= -1

1 + tan²(x) = sec²(x)

sec²(x)-tan(x) = 1, not -1

false

C. -cos^2 theta-1= sin^2

cos²(x) + sin²(x) = 1

sin²(x) = 1-cos²(x), our 1 is positive not negative, so false

D. Cot^2 theta - csc^2 theta=-1

cot²(x) + 1 = csc²(x)

isolating 1

1 = csc²(x) - cot²(x)

multiplying both sides by -1

-1 = cot²(x) - csc²(x)

TRUE

3 0
3 years ago
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