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tekilochka [14]

3 years ago

15

Rewrite the following exponential expressions as equivalent radical expressions. If the number is rational, write it without rad

icals or exponents.
(1/8)^5/3

1 answer:

alina1380 [7]3 years ago

7 0

Answer:

1/32

Step-by-step explanation:

simplify

1/[(2×2×2)^⅓]^5

= [1/(2³)^⅓]^5

= [1/2]^5

= [1/(2×2×2×2×2)]

= [1/32]

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Write 0.00605 in scientific notation.

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The answer to that is 3 jumps. Since the move of the decimal made the number larger than it really is, it will take a negative value to put it back where it belongs.

So A) is the correct answer. It is in scientific notation and if you move the decimal back 3 place you get the number you stared with.

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What are the possible values of x for the following functions?

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What can a denominator never be?

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to figure this out, we need to set (x-1)(x-2)=0

now we split and solve.

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ALL REAL NUMBERS not equal to 1 or 2

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In July, 3/5 of the 160 babies born in an area hospital were girls. Seventy of the babies weighed 8 pounds or more. Fifty boys w

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3 years ago

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Arlecino [84]

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5 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