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evablogger [386]
3 years ago
11

How many miles does a runner travel in a five-kilometer race?

Mathematics
1 answer:
ollegr [7]3 years ago
4 0
Since one kilometer is equal to about 0.621 miles, a runner travels 3.105 miles in a five-kilometer race.
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Area of the figure: A)128 B)136 C)153 C)258
GarryVolchara [31]

Answer: 153

Step-by-step explanation:

area of rectangle= 16 x 3= 48

area of trapezium= 1/2(a+b)h

=1/2(16+5)x 10

=105

105+48= <u>153 in^2</u>

3 0
3 years ago
8/(-4)+7<br><br><br><br> help LMFO <br> and explain
Vilka [71]

Answer:

5

Step-by-step explanation:

reminder of rules for division

• If signs are the same then positive result

• If signs are different the negative result

\frac{8}{-4} + 7 ( signs of division are different )

= - 2 + 7

= 5

7 0
3 years ago
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An online store store uses the following model to sell sneakers to its customers:
r-ruslan [8.4K]

Answer:

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Step-by-step explanation:

4 0
2 years ago
Please help me on my discussion topic.i'll give you 20pts 
Crazy boy [7]

Artists such as Al Held, Bridget Riley etc are known for their abstract artwork or geometric artwork. Riley is famous for her optical artwork.

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Besides optical art, other abstract art forms are also famous. These imbibe a strong imagination and a sense of creativity in its viewers. This art form puts different colors, shapes, and textures together to create a finished piece that represents something in particular.

Yes, I appreciate this kind of art form because these forms are genuine and helps us understand the art patterns. Unlike the traditional form, these forms explore the relationships of forms/shapes and colors. Moreover, art always reflects culture. Therefore, abstract art is important because it is reflecting a culture that has been moving since last 2000 years.

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3 years ago
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HELP HOW DO I SOLVE THE FOLLOWING? PLEASE HELP
levacccp [35]
f(x) = \sqrt[3]{2x - 1} \\f(14) = \sqrt[3]{2(14) - 1} \\f(14) = \sqrt[3]{28 - 1} \\f(14) = \sqrt[3]{27} \\f(14) = 3

f(x) = \sqrt[3]{2x - 1} \\f(-62) = \sqrt[3]{2(-62) - 1} \\f(-62) = \sqrt[3]{-124 - 1} \\f(-62) = \sqrt[3]{-125} \\f(-62) = -5

f(x) = \sqrt[3]{2x - 1} \\f(20) = \sqrt[3]{2(20) - 1} \\f(20) = \sqrt[3]{40 - 1} \\f(20) = \sqrt[3]{39}
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3 years ago
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