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3 years ago

A parabolic mirror to be used in a

Mathematics

1 answer:

jok3333 [9.3K]3 years ago

8 0

I’m sorry if I’m wrong I’d stay safe and let someone anwser tell u but I think 20

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The four-sided geometric figure pictured is called a parallelogram. One feature of parallelograms is that opposite sides have eq

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B is correct answer.

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Find mzP A 50 B. 60 C. 70 D. 120

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Answer:

d.120

Step-by-step explanation:

p=70

m=120

z=120

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3 years ago

The two dot plots show the number of goals per game. The data for each is taken from a random sample within a soccer league over

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first blank is 3

second blank is 2.5

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Step-by-step explanation:

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2 years ago

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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

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