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guajiro [1.7K]
3 years ago
6

X 20) Becca has 3 times more games than Susan has. Write an algebraic

Mathematics
1 answer:
vova2212 [387]3 years ago
7 0

Answer:

3*x

Step-by-step explanation:

If Becca has 3 times the amount of games as Susan then you would multiply the number of games Susan has by 3. So we can use X (or another letter) to represent the number of games Susan has. Giving us 3*x  

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Answer: a=4

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Step-by-step explanation:

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What is 1 times 1? Please answer because I need this due in 15 minutes
Sonbull [250]

Answer:

1

Step-by-step explanation:

1*1=1

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A certain forest covers an area of
iVinArrow [24]

Answer:

2598 square kilometers

Step-by-step explanation:

Hello

Step 1

year one

using a rule of three is possible to find how much is 8.75 od 4500 km2

Let

if

4500 km2  ⇒ 100$

x?km2         ⇒8.75

do the relation and isolate x

4500:100\\x:8.75\\\frac{4500}{100}=\frac{x}{8.75}\\x=\frac{4500*8.75}{100} \\x=393.75\\

at the end of the year one, the area will be

4500-393.75=4106.25

this will be the initial area for the year 2.

Step 2

repite the step 1 with area initial =4106.25 km2

4106.25 km2  ⇒ 100$

x?km2         ⇒8.75

do the relation and isolate x

4106.25:100\\x:8.75\\\frac{4106.25}{100}=\frac{x}{8.75}\\x=\frac{4106.25*8.75}{100} \\x=359.29\\

at the end of the year 2, the area will be

4106-359.29=3746.70

this will be the initial area for the year 3.

Step 3

repite the step 1 with area initial =4106.25 km2

3746.70 km2  ⇒ 100$

x?km2         ⇒8.75

do the relation and isolate x

3746.70:100\\x:8.75\\\frac{3746.70}{100}=\frac{x}{8.75}\\x=\frac{3746.7*8.75}{100} \\x=327.83\\

at the end of the year 3, the area will be

3746.70-327.83=3419.09

this will be the initial area for the year  4.

Step 4

year four

repite the step 1 with area initial =3419.09 km2

3419.09 km2  ⇒ 100$

x?km2         ⇒8.75

do the relation and isolate x

3419.09:100\\x:8.75\\\frac{3419.09}{100}=\frac{x}{8.75}\\x=\frac{3419.09*8.75}{100} \\x=299.17\\

at the end of the year 4, the area will be

3419.09-299.173=3119.82

this will be the initial area for the year  5.

Step 5

year five

repite the step 1 with area initial =3119.82 km2

3119.82 km2  ⇒ 100$

x?km2         ⇒8.75

do the relation and isolate x

3119.82:100\\x:8.75\\\frac{3119.82}{100}=\frac{x}{8.75}\\x=\frac{3119.82*8.75}{100} \\x=272.99\\

at the end of the year 5, the area will be

3119.82-272.99=2846.92

this will be the initial area for the year  6.

Step 6

year six

repite the step 1 with area initial =2846.92km2

2846.92 km2  ⇒ 100$

x?km2         ⇒8.75

do the relation and isolate x

2846.92:100\\x:8.75\\\frac{2846.92}{100}=\frac{x}{8.75}\\x=\frac{2846.92*8.75}{100} \\x=249.10\\

at the end of the year six, the area will be

2846.92-249.10=2597.82 square kilometers

Have a great day.

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Simplify the expression r^-3 s^5 t^2/r^2 st^-2
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Hello here is a solution : 

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