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3 years ago

8

Solve for brainliest!

1 answer:

balandron [24]3 years ago

8 0

Answer:

$9.50

Step-by-step explanation:

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The circumference covered by a radar signal can be represented by the expression 400π miles. What is the approximate length of t

goldfiish [28.3K]

Answer:

127 if it says to use 3.14, 200 if it does not ask to use 3.14

Step-by-step explanation:

C= $2\pi r$

Substitute C for 400

400=2*r

400=2r

divided both sides by 2

Answer is 200

8 0

3 years ago

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Stans son is 57 inches tall. what is his height in feet and inches?

Elenna [48]

Remember that one foot is 12 inches.
So a 57 inch person would be 4 feet, 9 inches.

Good luck!

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3 years ago

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True or False: It is possible for two 80 degree angles to form a linear pair

jekas [21]

The answer is false sorry it is true no it is false sorry don,t know

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3 years ago

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Three if the angles of a quadrilateral are 148, 104 and 36 whats the size of the 4th angle

vesna_86 [32]

The size of the 4th angle is 72.

Given:
1st angle = 148
2nd angle = 104
3rd angle = 36
shape = quadrilateral.

A quadrilateral is a shape that has 4 sides or edges, 4 corners or vertices.
It goes to show that it must also have 4 angles. These four angles must give the sum of 360°.

Since the first three angles are given, we can get the fourth angle by deducting the sum of the three angles from 360°

4th angle = 360° - 148° - 104° - 36°
= 360° - 288°
4th angle = 72°

3 0

3 years ago

Explain me PLEASEEEE!!!

kirza4 [7]

Answer:

$\frac{(a^{2})^{3}.a^{-3}}{a^{10}}=\frac{a^{6}.a^{-3}}{a^{10}}=\frac{a^{3}}{a^{10}}=a^{-7}=\frac{1}{a^{7}}$

Step-by-step explanation:

Let us revise the properties of exponents

$a^{m}.a^{n}=a^{m+n}$

$\frac{a^{m}}{a^{n}}=a^{m-n}$

$(a^{m})^{n}=a^{m.n}$

$a^{-m}=\frac{1}{a^{m} }$

Let us use these properties to solve the question

→ By using the 3rd property above

∵ $(a^{2})^{3}=a^{2.3}=a^{6}$

∴ $\frac{(a^{2})^{3}.a^{-3}}{a^{10}}=\frac{a^{6}.a^{-3}}{a^{10} }$

→ By using the 1st property above

∵ $a^{6}.a^{-3}=a^{6+-3}=a^{6-3}=a^{3}$

∴ $\frac{(a^{2})^{3}.a^{-3}}{a^{10}}=\frac{a^{6}.a^{-3}}{a^{10}}=\frac{a^{3}}{a^{10}}$

→ By using the 2nd property above

∵ $\frac{a^{3}}{a^{10}}=a^{3-10}=a^{-7}$

∴ $\frac{(a^{2})^{3}.a^{-3}}{a^{10}}=\frac{a^{6}.a^{-3}}{a^{10}}=\frac{a^{3}}{a^{10}}=a^{-7}$

→ By using the 4th property above

∵ $a^{-7}=\frac{1}{a^{7}}$

∴ $\frac{(a^{2})^{3}.a^{-3}}{a^{10}}=\frac{a^{6}.a^{-3}}{a^{10}}=\frac{a^{3}}{a^{10}}=a^{-7}=\frac{1}{a^{7}}$

8 0

3 years ago