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3 years ago

15

Leon throws a ball off a cliff. The graph represents a relation that models the ball’s height, h, in feet over time, t, in secon

ds.
What is the domain?

{t| 0 ≤ t ≤ 3}
{t| 0 ≤ t ≤ 48}
{t| 0 ≤ t ≤ 64}
{t| all real numbers}

2 answers:

erik [133]3 years ago

8 0

Answer: The answer is A.

Step-by-step explanation:

otez555 [7]3 years ago

4 0

Answer:

The correct option is 1.

Step-by-step explanation:

The given graph represents a relation that models the ball’s height, h, in feet over time, t, in seconds.

The value of height and time can not be negative, there the values of h and t must be greater than 0.

The set of all possible inputs is called domain. In the graph x axis represent the domain and y-axis represent the range.

The graph is defined from t=0 to t=3.

$Domain=\{t|0\leq t\leq3\}$

Therefore option 1 is correct.

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Step-by-step explanation:

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Reason -

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Determine what shape is formed for the given coordinates for ABCD, and then find the perimeter and area as an exact value and ro

Helga [31]

Answer:

Part 1) The shape is a trapezoid

Part 2) The perimeter is $25(4+\sqrt{2})\ units$ or approximately $135.4\ units$

Part 3) The area is $937.5\ units^2$

Step-by-step explanation:

step 1

Plot the figure to better understand the problem

we have

A(-28,2),B(-21,-22),C(27,-8),D(-4,9)

using a graphing tool

The shape is a trapezoid

see the attached figure

step 2

Find the perimeter

we know that

The perimeter of the trapezoid is equal to

$P=AB+BC+CD+AD$

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Find the distance AB

we have

A(-28,2),B(-21,-22)

substitute in the formula

$d=\sqrt{(-22-2)^{2}+(-21+28)^{2}}$

$d=\sqrt{(-24)^{2}+(7)^{2}}$

$d=\sqrt{625}$

$d_A_B=25\ units$

Find the distance BC

we have

B(-21,-22),C(27,-8)

substitute in the formula

$d=\sqrt{(-8+22)^{2}+(27+21)^{2}}$

$d=\sqrt{(14)^{2}+(48)^{2}}$

$d=\sqrt{2,500}$

$d_B_C=50\ units$

Find the distance CD

we have

C(27,-8),D(-4,9)

substitute in the formula

$d=\sqrt{(9+8)^{2}+(-4-27)^{2}}$

$d=\sqrt{(17)^{2}+(-31)^{2}}$

$d=\sqrt{1,250}$

$d_C_D=25\sqrt{2}\ units$

Find the distance AD

we have

A(-28,2),D(-4,9)

substitute in the formula

$d=\sqrt{(9-2)^{2}+(-4+28)^{2}}$

$d=\sqrt{(7)^{2}+(24)^{2}}$

$d=\sqrt{625}$

$d_A_D=25\ units$

Find the perimeter

$P=25+50+25\sqrt{2}+25$

$P=(100+25\sqrt{2})\ units$

simplify

$P=25(4+\sqrt{2})\ units$ ----> exact value

$P=135.4\ units$

therefore

The perimeter is $25(4+\sqrt{2})\ units$ or approximately $135.4\ units$

step 3

Find the area

The area of trapezoid is equal to

$A=\frac{1}{2}[BC+AD]AB$

substitute the given values

$A=\frac{1}{2}[50+25]25=937.5\ units^2$

4 0

3 years ago

Pls help me with this no links please and Thank you

den301095 [7]

The top graph is correct because an x=# will create a vertical line at that #

4 0

3 years ago

Read 2 more answers