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Serjik [45]
3 years ago
11

Please help! f(x)=x^2. What is g(x)

Mathematics
1 answer:
lorasvet [3.4K]3 years ago
4 0

Answer:

D

Step-by-step explanation:

The standard form for a parabola is ax^2+bx+c. In this equation, b=0 and c=0. For g(x), you would plug in your y-value that was given (which is 4) and you would plug in your x-value that that given (which is 1). Use PEMDAS. The correct answer is the one that is true.

A. g(x) = (4x)^2

4 = (4(1))^2

4 = 16. (not true)

B. g(x) = 1/4 x^2

4 = 1/4 (1)^2

4 = 1/4 (not true)

C. g(x) = 16x^2

4 = 16(1)^2

4 = 16 (not true)

D. g(x) = 4x^2

4 = 4(1)^2

4 = 4. (true --> correct answer)

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1. whats the midpoint between (7,3) and (-3,-1)
Elanso [62]

Answer:

(2, 1 )

Step-by-step explanation:

Given endpoints (x₁, y₁ ) and (x₂, y₂ ) then the midpoint is

( \frac{x_{1}+x_{2}  }{2}, \frac{y_{1}+y_{2}  }{2} )

Here (x₁, y₁ ) = (7, 3) and (x₂, y₂ ) = (- 3, - 1) , then

midpoint = ( \frac{7-3}{2}, \frac{3-1}{2} ) = ( \frac{4}{2}, \frac{2}{2} ) = (2, 1 )

5 0
3 years ago
Whats the answer to 26.02 x 2.006
finlep [7]

Answer:

52.19612

Step-by-step explanation:

hope this helps

8 0
4 years ago
What is the volume of the cone?
mr_godi [17]

1/3 (3.14) (2*2) (4)


8 0
3 years ago
A helicopter flew north 325 meters and then flew east 500 meters. How far is the helicopter from its starting point?
kati45 [8]

Answer:

596.34m approx

Step-by-step explanation:

Given data

Let the Starting point be x

A helicopter flew north 325 meters from x

Then flew east 500 meters

Let us apply the Pythagoras theorem to solve for the resultant which is the distance from the starting position

x^2= 325^2+500^2

x^2=105625+250000

x^2= 355625

x= √355625

x=596.34m

Hence the distance from the starting point is 596.34m approx

8 0
3 years ago
A child lets go of a balloon that rises at a constant rate. 5 seconds after it was released, the balloon is at a height of 16 fe
Mama L [17]

Answer:

1. h = 2.4t + 4

2. 4 feet

3. 220 feet

Step-by-step explanation:

1. Write a linear model for the height, h, of the balloon as a function of the number of seconds, s that it has been raising.

Since the balloon rises at a constant rate, we find this rate by using the initial and final values of height and time at 5 seconds and 20 seconds respectively which are 12 feet and 52 feet respectively.

So, rate = gradient of line

= change in height/change  in time

= (52 ft - 16 ft)/(20 s - 5 s)

= 36 ft/15 s

= 2.4 ft/s

Now the equation of the line which shows the height is gotten from

(h - h')/(t - t') = rate

Using h'= 16 feet and t' = 5 s, we have

(h - 16)/(t - 5) = 2.4

h - 16 = 2.4(t - 5)

h - 16 = 2.4t - 12

h = 2.4t - 12 + 16

h = 2.4t + 4

where h is the height of the balloon above the ground and t is the time spent in the air in seconds.

2. What was the height of the balloon initially before the child let it go?

We obtain the initial height of the balloon before the child let go at time, t = 0 the time before the child let go.

So, substituting t = 0 into the equation for h, we have

h = 2.4t + 4

h = 2.4(0) + 4

h = 0 + 4

h = 4 feet

So, the height of the balloon before the child let go is 4 feet above the ground.

3. Use your model to predict the height of the balloon after 90 seconds.

We insert t = 90 s into the equation for h. So,

h = 2.4t + 4

h = 2.4(90) + 4

h = 216 + 4

h = 220 feet

So, the height of the balloon after 90 s is 220 feet above the ground.

8 0
3 years ago
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