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Stella [2.4K]
3 years ago
13

13 ft :10 ft 18 ft Find the area of the trapezoid. [? ] square feet Hint: The formula for the area of a trapezoid is: (b17b2). h

​
Mathematics
1 answer:
Liono4ka [1.6K]3 years ago
3 0
The area is 29473
i just need points
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Suppose an arrow is shot upward on the moon with a velocity of 44 m/s, then its height in meters after tt seconds is given by h(
andreev551 [17]

Answer:

a. 38.19m/s

b. 38.605m/s

c. 38.937m/s

d. 39.0117m/s

e. 39.01917m/s

Step-by-step explanation:

The average velocity is defined as the relationship between the displacement that a body made and the total time it took to perform it. Mathematically is given by the next formula:

v_a_v_g = \frac{\Delta x}{\Delta t} =\frac{x_f-x_i}{t_f-t_i}

Where:

x_f=Final\hspace{3}distance\hspace{3}traveled\\x_i=Initial\hspace{3}distance\hspace{3}traveled\\t_f=Final\hspace{3}time\hspace{3}interval\\t_i=Initial\hspace{3}time\hspace{3}interval

a. Let's find h(3) and h(4) using the data provided by the problem:

h(3)=44(3)-0.83(3^2)=124.53=x_i\\h(4)=44(4)-0.83(4^2)=162.72=x_f

The average velocity over the interval [3, 4] is :

v_a_v_g=\frac{162.72-124.53}{4-3} =38.19m/s

b. Let's find h(3.5) using the data provided by the problem:

h(3.5)=44(3.5)-0.83(3.5^2)=143.8325=x_f

The average velocity over the interval [3, 3.5] is :

v_a_v_g=\frac{143.8325-124.53}{3.5-3} =38.605m/s

c. Let's find h(3.1) using the data provided by the problem:

h(3.1)=44(3.1)-0.83(3.1^2)=128.4237=x_f

The average velocity over the interval [3, 3.1] is :

v_a_v_g=\frac{128.4237-124.53}{3.1-3} =38.937m/s

d. Let's find h(3.01) using the data provided by the problem:

h(3.1)=44(3.01)-0.83(3.01^2)=124.920117=x_f

The average velocity over the interval [3, 3.01] is :

v_a_v_g=\frac{124.920117-124.53}{3.01-3} =39.0117m/s

e. Let's find h(3.001) using the data provided by the problem:

h(3.001)=44(3.001)-0.83(3.001^2)=124.5690192=x_f

v_a_v_g=\frac{124.5690192-124.53}{3.001-3} =39.01917m/s

7 0
3 years ago
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