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4 years ago

10

which power can you write to represent the volume of a cube that is 1/3 inches. write the power as an expression with a base and

an exponent

2 answers:

otez555 [7]4 years ago

8 0

Base: 1/3
Exponent: 3
Hope this helped

Kipish [7]4 years ago

3 0

Answer: The volume of a cube that is 1/3 inches is represented by $(\frac{1}{3})^3$ .

Step-by-step explanation:

Given: The side length of cube = $\frac{1}{3}$ inches

The volume of cube with side length 'a' is given by :-

Volume= $a^3$ , where base=a and power =3

Therefore, the volume of cube with side length $\frac{1}{3}$ inches is given by :-

Volume= $(\frac{1}{3})^3$

Here, Base= $\frac{1}{3}$

Power=3

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3 years ago

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3 years ago

Which point on the graph of g(x)=(1/5)^x? HELPP

Cloud [144]

Answer:

(-1,5) and $(3, \frac{1}{125})$ are points on the graph

Step-by-step explanation:

Given

$g(x) = \frac{1}{5}^x$

Required

Determine which point in on the graph

To get which of point A to D is on the graph, we have to plug in their values in the given expression using the format; (x,g(x))

A. (-1,5)

x = -1

Substitute -1 for x in $g(x) = \frac{1}{5}^x$

$g(x) = \frac{1}{5}^{-1}$

Convert to index form

$g(x) = 1/(\frac{1}{5})$

Change / to *

$g(x) = 1*(\frac{5}{1})$

$g(x) = 5$

This satisfies (-1,5)

<em>Hence, (-1,5) is on the graph</em>

<em></em>

B. (1,0)

x = 1

Substitute 1 for x

$g(x) = \frac{1}{5}^x$

$g(x) = \frac{1}{5}^1$

$g(x) = \frac{1}{5}$

<em>(1,0) is not on the graph because g(x) is not equal to 0</em>

C. $(3, \frac{1}{125})$

x = 3

Substitute 3 for x

$g(x) = \frac{1}{5}^x$

$g(x) = \frac{1}{5}^3$

Apply law of indices

$g(x) = \frac{1}{5} * \frac{1}{5} * \frac{1}{5}$

$g(x) = \frac{1}{125}$

This satisfies $(3, \frac{1}{125})$

<em>Hence, </em> $(3, \frac{1}{125})$ <em> is on the graph</em>

<em></em>

D. $(-2, \frac{1}{25})$

x = -2

Substitute -2 for x

$g(x) = \frac{1}{5}^x$

$g(x) = \frac{1}{5}^{-2}$

Convert to index form

$g(x) = 1/(\frac{1}{5}^2)$

$g(x) = 1/(\frac{1}{5}*\frac{1}{5})$

$g(x) = 1/(\frac{1}{25})$

Change / to *

$g(x) = 1*(\frac{25}{1})$

$g(x) = 25$

This does not satisfy $(-2, \frac{1}{25})$

<em>Hence, </em> $(-2, \frac{1}{25})$ <em> is not on the graph</em>

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4 years ago