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kari74 [83]
3 years ago
10

Please help I need it please

Mathematics
1 answer:
Sergio039 [100]3 years ago
5 0

Answer:

30%

Step-by-step explanation:

7000/10000 = 7/10 = .7 = 70%

100%-70% = 30%

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There are 21 nickles in the jar.
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Erin wants to buy a set of paintbrushes. Store A sells a set for $25 and offers a 10% discount. Store B sells a set for $27 and
Deffense [45]
A: 25x(1-0.1)=22.5
B: 27x(1-0.2)=21.6
C: 26x(1-0.15)=22.1
Answer: Store B (lowest price), store C and store A (highest price).
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3 years ago
What is the area of a regular pentagon with a perimeter of 115cm?
shusha [124]
About 480.4 or 480.4262466 if you need the whole thing
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4 years ago
Ann increased the quantities of all the ingredients in a recipe by 60% She used 80 grams (g) of cheese How much cheese did the r
Delvig [45]

Answer:g = 50 g

Step-by-step explanation:

She used 60% more than required ⇒ she used (1+0.60)=1.60 times what was required.

x = grams required by the recipe

80 g = 1.60x

x = (80/1.60) g = 50 g

3 0
3 years ago
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.

8 0
3 years ago
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