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xxTIMURxx [149]
3 years ago
10

Which equation represents a parabola with a vertex at the origin and a focus at (0,-1)

Mathematics
1 answer:
tensa zangetsu [6.8K]3 years ago
5 0

We are given parabola with vertex at origin (0,0) and given focus (0,-1).

We know vertex is given by (h,k).

Therefore, h=0 and k=0.

Formula for focus is (h, k + p).

On comapring with given focus

<em>k+p = -1.</em>

Plugging value of k=0 in above equation we get

0 +p =-1.

p = -1.

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

Plugging values of h, k and p in parabola equation, we get

4(-1) (y-0) = (x-0)^2

-4y = x^2

Dividing both sides by -4, we get

y=-\frac{1}{4}x^2.

<h3>Therefore, y=-\frac{1}{4}x^2 equation represents a parabola with a vertex at the origin and a focus at (0,-1).</h3>
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Step-by-step explanation:

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6 0
3 years ago
What are the solutions to the absolute value inequality |x − 70| ≤ 3? Remember, the inequality can be written as −3 ≤ x − 70 ≤ 3
ikadub [295]

Answer:

solutions to the absolute value inequality is 67\leq x\leq 73

Step-by-step explanation:

To find the solution of x absolute value of  |x − 70| ≤ 3 will be written in the form of interval because the given fraction (x-70) is less than and equal to 3.

-3\leq(x-70)\leq 3

Now we add 70 on every part of the inequality.

-3+70\leq (x-70)+70\leq 3+70

67\leq x\leq 73

So the solution to the absolute value inequality is 67\leq x\leq 73.


5 0
3 years ago
Read 2 more answers
a rectangular table is six times as long as it is wide. if the area is 96 ft^2, find the length and the width of the table
Deffense [45]
I found this for you. It even tells you how the answer was achieved.



7 0
4 years ago
Which equations represent the asymptotes of the hyperbola (x-1)^2/36-(y-2)^2/64=1 ?
allochka39001 [22]

Answer:

4x+3y=10 and  4x-3y=-2

Step-by-step explanation:

The asymptote equation of a translate hyperbola with equation \frac{(x-h)^2}{b^2}-\frac{(y-k)^2}{a^2}=1 is y=\pm\frac{b}{a}(x-h)+k

The given hyperbola has equation: \frac{(x-1)^2}{36}-\frac{(y-2)^2}{64}=1

Or \frac{(x-1)^2}{6^2}-\frac{(y-2)^2}{8^2}=1

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

We have  a=6 and b=8, h=1 and k=2.

We substitute this values to get:

y=\pm\frac{6}{8}(x-1)+2

y=\pm\frac{4}{3}(x-1)+2

3y=\pm4(x-1)+6

We split the plus or minus sign to get:

3y=4x-4+6

3y=4x+2

3y-4x=2

Or

3y=-4(x-1)+6

3y=-4x+4+6

3y+4x=10

5 0
4 years ago
(y = 4x + 3<br> y = -x - 2<br> Please help
Gnom [1K]
I think that it would be 3x + 1.
4 0
3 years ago
Read 2 more answers
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