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galben [10]
3 years ago
14

Round 22.5240 to the nearest thousandth

Mathematics
1 answer:
shutvik [7]3 years ago
5 0
22.5240 rounded to the nearest thousandth is 22.524. For this decimal number to be rounded up rather than down, the original number would have had to been 22.525 or higher.
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8X 1 + 5 X 1/100 + 9 X 1/1000
USPshnik [31]
I hope this helps you

7 0
3 years ago
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If the function h is defined by h(x)=<img src="https://tex.z-dn.net/?f=x%5E%7B2%7D" id="TexFormula1" title="x^{2}" alt="x^{2}" a
harkovskaia [24]

Given:

The function is:

h(x)=x^2-3x+5

To find:

The value of h(2x+1).

Solution:

We have,

h(x)=x^2-3x+5

Putting x=2x+1, we get

h(2x+1)=(2x+1)^2-3(2x+1)+5

h(2x+1)=(2x)^2+2(2x)(1)+(1)^2-3(2x)-3(1)+5

h(2x+1)=4x^2+4x+1-6x-3+5

On combining like terms, we get

h(2x+1)=4x^2+(4x-6x)+(1-3+5)

h(2x+1)=4x^2-2x+3

Therefore, the required function is h(2x+1)=4x^2-2x+3.

3 0
3 years ago
Which of the following is a metric unit of measure that measures length and distance? Foot Meter Area Yard
Liula [17]

Area is not a unit of measure. Foot and Yard are US customary units. Meter is the correct answer.

3 0
3 years ago
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The equation below can be used to solve which of the following word problems? 2x + 4y = 18
densk [106]
One of the 4 options prolly D
7 0
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>

8 0
3 years ago
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