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vitfil [10]
3 years ago
6

Question 1 of 10

Mathematics
2 answers:
jonny [76]3 years ago
8 0

Answer:

A.

Step-by-step explanation:

Just did it

kvasek [131]3 years ago
4 0

Given:

A parabola opens up or down and its vertex is at the point (h, v).

To find:

The equation of the parabola.

Solution:

A parabola opens up or down. It means it is a vertical parabola.

The vertex form of a vertical parabola is

y=a(x-h)^2+k             ...(i)

Where, a is constant and (h,k) is vertex.

The vertex of given parabola is at point (h,v). Putting h=h and k=v in (i), we get

y=a(x-h)^2+v

The required equation of the parabola is y=a(x-h)^2+v.

Therefore, the correct option is A.

You might be interested in
Pythagorean theorem
marta [7]

Answer:

x = 8

Step-by-step explanation:

a^{2} +b^{2} =c^{2} \\ so 8^{2} +b^{2} =\sqrt{80} ^{2}

rearrange the formula:

\sqrt{80} ^{2} - 8^{2} = b^{2}

80 - 64 = b^2

b^2 = 16\\b  = \sqrt{16}

\sqrt{16}  = 4

x = 4+4 = 8

7 0
3 years ago
PLEASE HELP ME UwU
klasskru [66]
The answer is obviously 100 that was so easy
3 0
3 years ago
Read 2 more answers
I need an answer ASAP pls
tino4ka555 [31]

Answer:

Age 42 salary 30,000

Step-by-step explanation:

He had to restart his medical training, therefore, making him have a lower salary regardless of his age, making this point an obvious outlier

4 0
3 years ago
Please anyone help me
sineoko [7]
Hey There!!!!


The Answer Is > 240 x 3 = 720 so 720 is your answer!!


Please Mark Me Brainliest :) :) Annierosehodge (: (:
6 0
3 years ago
Read 2 more answers
For the following integral, give a power or simple exponential function that if integrated on a similar infinite domain will hav
zhannawk [14.2K]

Answer:

hello your question is incomplete attached below is the complete question

answer :

for  I1 =     \frac{1}{x^2} + \frac{4}{x^3}    The integral converges

for  I2 =    \frac{2x + 6}{2(x^2 + 6x + 4 )}  The integral diverges

for  I3 =  \frac{1}{x^2}  The integral converges

for  I4 = 1  The integral diverges

Step-by-step explanation:

The similar integrands and the prediction ( conclusion )

for  I1 =     \frac{1}{x^2} + \frac{4}{x^3}    The integral converges

for  I2 =    \frac{2x + 6}{2(x^2 + 6x + 4 )}  The integral diverges

for  I3 =  \frac{1}{x^2}  The integral converges

for  I4 = 1  The integral diverges

attached below is a detailed solution

5 0
2 years ago
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