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snow_lady [41]
3 years ago
10

Find the percent decrease,and round to the nearest percent. from 64 photos to 21 photos

Mathematics
1 answer:
Tom [10]3 years ago
6 0
64-21=43
43/64=.71

71%
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How many 1/5 s make 14/10
Sliva [168]

Answer:

7

Step-by-step explanation:

\frac{14}{10}  \div  \frac{1}{5}  = 7

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2 years ago
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)) Omar used 28 centimeters of tape to wrap 4 presents. How much tape will Omar need in
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Answer:

He would need 42 cm *or 14cm

Step-by-step explanation:

28/4 is 7

that means that he used 7 cm of tape per present

6 x 7 is 42

that means that he would need 42 cm of tape

4 to 28

6 to 42

*If he already has 4 presents done and he needs to 2 more present then he would need 14 cm, 2 x 7 is 14

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3 years ago
Kayla has a piece of blue ribbon that is 18.35 inches long and a piece of
tankabanditka [31]

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1.85

Step-by-step explanation:

20.2-18.35=1.85

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Please help with b) and c)
rewona [7]
For b it is A,M and T
For C it is 0%
6 0
2 years ago
The cost of renting a bus for a field trip was split evenly among 20 students. At the last minute, 10 more students joined the t
dexar [7]

By solving a linear equation, we will see that the total cost for renting the bus is $90.

<h3>What was the total cost of renting the bus, in dollars?</h3>

Let's say that the total cost is C.

When there are 20 students, each student should pay:

p = C/20

When the other 10 students are added (for a total of 30) each student pays:

p' = C/30.

We know that the cost for each of the original 20 students decreased by $1.50, so:

p' = p - $1.50

Then we have 3 equations to work with:

p = C/20

p' = C/30.

p' = p - $1.50

Now we can replace the first and second equations into the third one:

C/30 = C/20 - $1.50

Now we can solve this linear equation for C:

C/20 - C/30 = $1.50

C*( 1/20 - 1/30) = $1.50

C*(30/600 - 20/600) = $1.50

C*(10/600) = $1.50

C*(1/60) = $1.50

C = 60*$1.50 = $90

So the total cost for renting the bus is $90.

If you want to learn more about linear equations:

brainly.com/question/1884491

#SPJ1

4 0
2 years ago
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