Can someone help me with a question!?!?

Mathematics

1 answer:

Len [333]3 years ago

8 0

If the parent graph f(x) = x² is changed to f(x) = 2x², the vertex of the parabola will still remain (0, 0) because whenever the equation of a parabola is in the form y = ax², the vertex will always be (0, 0).

Now if <em>a</em> is a big number, the parabola will become narrower.

So it will stretch vertically and become narrower.

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A father’s age now is three times the age that his son was 4 years ago. In

Mice21 [21]

Let ‘s’ be the son’s age 12 years ago.
Let ‘f’ be the father’s current age.

4 years ago, the son was:

s-4

So, his father is currently:

3(s-4)

=

3s-12

Therefore:

f = 3s-12

In twelve years, the son will be:

s+12

And the father will be:

f+12

This can also be written as:

3s-12+12 as the fathers younger age would be f = 3s+12

=

3s

So, we know that s+12 is half the fathers current age, meaning the father is currently 2(s+12) which is equivalent to 2s+24. Also, we know that the father is currently 3 times the sons age 12 years ago, so 3s (proven by the calculations we made above). Therefore, 2s+24=3s which means 24=s. We can then substitute this, and we will receive 24+12 = 36

Son’s current age: 36

We then substitute the son’s age 12 years ago into 2s+24 to give us the father’s age.

2(24)+24 = 72

Father’s current age: 72

7 0

1 year ago

A track-and-field athlete releases a javelin. The height of the javelin as a function of time is shown on the graph below. Use t

Georgia [21]

Part 1:

For this case we must see in the graph the axis of symmetry of the given parabola.
We have then that the axis of symmetry is the vertical line t = 2.
Answer:
The height of the javelin above the ground is symmetric about the line t = 2 seconds:

Part 2:

For this case, we must see the time t for which the javelin reaches a height of 20 feet for the first time.
We then have that when evaluating t = 1, the function is h (1) = 20. To do this, just look at the graph.
Then, we must observe the moment when it returns to be 20 feet above the ground.
For this, observing the graph we see that:
h (3) = 20 feet
Therefore, a height of 20 feet is again reached in 3 seconds.
Answer:
The javelin is 20 feet above the ground for the first time at t = 1 second and again at t = 3 seconds

5 0

3 years ago

the diameter of a cylindrical construction pipe is 4 ft. if the pipe is 19 ft long, what s it’s volume?

STALIN [3.7K]

Answer:

76 $\pi$ or 238.76 $units^{3}$