All categories

3 years ago

What is the volume of the trapezoidal right prism?

Mathematics

2 answers:

Sphinxa [80]3 years ago

8 0

A=(a+b)*h*1/2=(20+8)*7*1/2=49 m2
V=A*H=49*2=98 m3

Nikitich [7]3 years ago

8 0

V=areaofbase times height
height=2
base is a trapezoid
A=(1/2)(b1+b2)h
b1=8
b2=20
h=7
A=(1/2)(8+20)(7)
A=(1/2)(28)(7)
A=(14)(7)
A=98

V=BH
V=98(2)
V=196 m^3

You might be interested in

I need help pls !!!!!!

Rama09 [41]

(k-7) would be the equation

6 0

3 years ago

You put 5000 and a bank account that will give you 2% interest on the money in the account every year how much money will be in

ANEK [815]

Answer:

$7,429 I think (it has to be 20 characters ignore this part)

8 0

3 years ago

You are competing in a race. The table shows the times from last year's race. You want your time to be last year's median time w

olasank [31]

Answer:

28 ≤ x ≤40

Step-by-step explanation:

Arrange the data from the smallest to the largest

26, 27, 28,32, 32, 33, 35, 36, 38, 40,42,48

Identify the median value

26, 27, 28,32, 32,<u> 33, 35</u>, 36, 38, 40,42,48

Getting the median

(33+35)/2 = 68/2 = 34 minutes

You want your time to be last year's median time with an absolute deviation of at most 6 minutes

This mean ±6 minutes, hence this will be 34 ± 6

34+6=40 and 34-6=28

The inequality will be 28 ≤ x ≤40

7 0

3 years ago

Read 2 more answers

How do I solve question 6 through 8?<br> Solve for me

rewona [7]

The equations of the functions are y = -4(x + 1)^2 + 2, y = 2(x - 2)^2 + 1 and y = -(x - 1)^2 - 2

<h3>How to determine the functions?</h3>

A quadratic function is represented as:

y = a(x - h)^2 + k

<u>Question #6</u>

The vertex of the graph is

(h, k) = (-1, 2)

So, we have:

y = a(x + 1)^2 + 2

The graph pass through the f(0) = -2

So, we have:

-2 = a(0 + 1)^2 + 2

Evaluate the like terms

a = -4

Substitute a = -4 in y = a(x + 1)^2 + 2

y = -4(x + 1)^2 + 2

<u>Question #7</u>

The vertex of the graph is

(h, k) = (2, 1)

So, we have:

y = a(x - 2)^2 + 1

The graph pass through (1, 3)

So, we have:

3 = a(1 - 2)^2 + 1

Evaluate the like terms

a = 2

Substitute a = 2 in y = a(x - 2)^2 + 1

y = 2(x - 2)^2 + 1

<u>Question #8</u>

The vertex of the graph is

(h, k) = (1, -2)

So, we have:

y = a(x - 1)^2 - 2

The graph pass through (0, -3)

So, we have:

-3 = a(0 - 1)^2 - 2

Evaluate the like terms

a = -1

Substitute a = -1 in y = a(x - 1)^2 - 2

y = -(x - 1)^2 - 2

Hence, the equations of the functions are y = -4(x + 1)^2 + 2, y = 2(x - 2)^2 + 1 and y = -(x - 1)^2 - 2