Which are equations? Check all that apply.

SOLVE THE QUESTION BELOW ASAP

qwelly [4]

Answer:

Part A) The graph in the attached figure (see the explanation)

Part B) 16 feet

Part C) see the explanation

Step-by-step explanation:

Part A) Graph the function

Let

h(t) ----> the height in feet of the ball above the ground

t -----> the time in seconds

we have

$h(t)=-16t^{2}+98$

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex is a maximum

To graph the parabola, find the vertex, the intercepts, and the axis of symmetry

Find the vertex

The function is written in vertex form

so

The vertex is the point (0,98)

Find the y-intercept

The y-intercept is the value of the function when the value of t is equal to zero

For t=0

$h(t)=-16(0)^{2}+98$

$h(0)=98$

The y-intercept is the point (0,98)

Find the t-intercepts

The t-intercepts are the values of t when the value of the function is equal to zero

For h(t)=0

$-16t^{2}+98=0$

$t^{2}=\frac{98}{16}$

square root both sides

$t=\pm\frac{\sqrt{98}}{4}$

$t=\pm7\frac{\sqrt{2}}{4}$

therefore

The t-intercepts are

$(-7\frac{\sqrt{2}}{4},0), (7\frac{\sqrt{2}}{4},0)$

$(-2.475,0), (2.475,0)$

Find the axis of symmetry

The equation of the axis of symmetry in a vertical parabola is equal to the x-coordinate of the vertex

so

$x=0$ ----> the y-axis

To graph the parabola, plot the given points and connect them

we have

The vertex is the point $(0,98)$

The y-intercept is the point $(0,98)$

The t-intercepts are $(-2.475,0), (2.475,0)$

The axis of symmetry is the y-axis

The graph in the attached figure

Part B) How far is the artifact fallen from the time t=0 to time t=1

we know that

For t=0

$h(t)=-16(0)^{2}+98$

$h(0)=98\ ft$

For t=1

$h(t)=-16(1)^{2}+98$

$h(1)=82\ ft$

Find the difference

$98\ ft-82\ ft=16\ ft$

Part C) Does the artifact fall the same distance from time t=1 to time t=2 as it does from the time t=0 to time t=1?

we know that

For t=1

$h(t)=-16(1)^{2}+98$

$h(1)=82\ ft$

For t=2

$h(t)=-16(2)^{2}+98$

$h(2)=34\ ft$

Find the difference

$82\ ft-34\ ft=48\ ft$

so

The artifact fall 48 feet from time t=1 to time t=2 and fall 16 feet from time t=0 to time t=1

therefore

The distance traveled from t=1 to t=2 is greater than the distance traveled from t=0 to t=1

8 0

3 years ago

For the following hypothesis test, determine the null and alternative hypotheses. Also, classify the hypothesis test as two-tail

bagirrra123 [75]

Answer:

Option B:

$H_0: \mu = 22.1$

$H_a: \mu \neq 22.1$

Classification:

The hypothesis test is Two-tailed.

Step-by-step explanation:

The mean length of imprisonment for motor-vehicle theft offenders in this country is 22.1 months.

This means that the null hypothesis is that the mean is of 22.1 months, that is:

$H_0: \mu = 22.1$

A hypothesis test is to be performed to determine whether the mean length of imprisonment for motor-vehicle theft offenders in this city differs from the national mean of 22.1 months.

At the alternate hypothesis, we test if this mean is different of 22.1, that is:

$H_a: \mu \neq 22.1$

Which means that the answer is given by option b).

Which of the following is the correct classification of the hypothesis test?

We test if the mean is different from a value, which means that the hypothesis test is Two-tailed.

7 0

3 years ago

What is the area of this figure?

Elanso [62]

First of all we can draw a parallel line to divide the figure into a triangle and a rectangle as shown in the figure. To find the area of our rectangle, remember that the area of a rectangle is length times width, so $A_{r} =lw$ . Since we know for our figure that the length and width of our rectangle are 13cm and 6cm respectively, lets replace those values in our formula to get its area:
$A _{r} =(13cm)(6cm)$
$A=78cm^{2}$
Similarly, the area of a triangle is one half times base times height, so $At=( \frac{1}{2})bh$ . Since we know that our base is 8cm and our height 6cm, lets replace those values in our equation to find the area of our triangle:
$A_{t} =( \frac{1}{2})(8cm)(6cm)$
$A_{t} =24cm^{2}$

Now the only thing left is add our areas:
$A_{total} =78cm^{2}+24cm^{2} =102cm^{2}$

We can conclude that the correct answer is A. 102

4 0

3 years ago

Linear inequalities What is the solution to: -1/3 a + 4 ≤ 0

kati45 [8]

Step-by-step explanation:

-1/3 a + 4 ≤ 0

3*(-1/3a) + 4*3 ≤ 0*3

-a + 12 ≤ 0

12 ≤ a

12 ⩾a

a⩾12

8 0

2 years ago

If x,y and z are positive integers and 3x=4y=7z,

Pani-rosa [81]

3x=4y=7z
so
GCF:
3 * 4 * 7 = 84

3x = 84
x = 28

4y = 84
y = 21

7z = 84
z = 12

least possible value for
x + y + z = 28 + 21 + 12 = 61

answer
(d) 61

5 0

3 years ago