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rewona [7]
2 years ago
4

The right square pyramid with side length 4 and height 5 is enlarged by a scale factor of 3. What is the volume of the enlarged

pyramid? Round your answer to the nearest cubic foot.
Mathematics
1 answer:
liberstina [14]2 years ago
5 0

Check the picture below.

\textit{volume of a pyramid}\\\\ V=\cfrac{Bh}{3} ~~ \begin{cases} B=\stackrel{base's}{area}\\ h=height\\[-0.5em] \hrulefill\\ B=\stackrel{(4)(3)\times (4)(3)}{144}\\ h=15 \end{cases}\implies V=\cfrac{(144)(15)}{3}\implies V=720

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On a piece of paper , graph f(x)=2^x then determine which answer choice matches the graph you drew.
amid [387]

Answer:

The graph in the attached figure

Step-by-step explanation:

we have

f(x)=2^x

This is a exponential function of the form

y=a(b^x)

where

a is the initial value or the y-intercept

b is the base of the exponential function

If b>1 then is a exponential growth function

If b<1 then is a exponential decay function

In this problem

The y-intercept is equal to

For x=0

f(x)=2^0=1

The y-intercept is the point (0,1)

so

a=1

b=2

The value of b is greater than 1

so

Is a growth function

To plot the graph create a table with different values of x and y

For x=-1

f(x)=2^-1=0.5

point (-1,0.5)

For x=1

f(x)=2^1=2

point (1,2)

For x=2

f(x)=2^2=4

point (2,4)

For x=3

f(x)=2^3=8

point (3,8)

For x=4

f(x)=2^4=16

point (4,16)

Plot the y-intercept and the other points and connect them to graph the exponential function

Note that as x increases the value of y increases (exponential growth function)

The graph in the attached figure

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4 years ago
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3 years ago
Select 3 ratios that are equivalent to 8:5
torisob [31]
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blagie [28]
SIMILARITIES
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3 0
4 years ago
Mike just hopped on the edge of a merry-go-round. What are his linear and angular speeds if the diameter of the merry-go-round i
dsp73

Step-by-step explanation:

Given that,

The diameter of the merry-go-round, d = 14 feet

Time taken, t = 6 seconds

Radius, r = 7 feet

The linear speed of the merry-go-round is given by :

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Also,

v=r\omega

Where

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So,

\omega=\dfrac{v}{r}\\\\\omega=\dfrac{7.33}{7}\\\\=1.04\ rad/s

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8 0
4 years ago
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