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aleksklad [387]
3 years ago
9

Find the diameter of a circle whose circumference is 18.85 m.

Mathematics
1 answer:
Inga [223]3 years ago
6 0
12 because  the formula for a circle is  pi r^2 (h) and u get the circumference  to get the diameter u just do it backwards

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Kayla squared a number x and added this result to -6.5. This gave her an answer of 42.5.
ladessa [460]

Answer:

7, -7

Step-by-step explanation:

x^2-6.5= 42.5 add 6.5 to both sides

x^2= 49

The square root of 49 is 7 and -7.

7 0
3 years ago
One angle of a triangle measures 36.5 degrees. What is the measure of the smaller angle?
Sunny_sXe [5.5K]

The final answer would depend in the type of triangle we are analyzing, however here are the possible outcomes:

1.) If it was a right triangle, 36.5 would be the smaller angle.

2.) It cannot be an equilateral triangle since all angles would be 60°.

3.) In a isosceles triangle, 36.5° would be the smaller, since the others would be 72°.

4.) In an scalene triangle it cannot be determined unless we had 2 angles since in that kind of triangle all angles can be different.

5.) In an acute triangle, 36.5° would be the smaller angle.

6.) In an obtuse triangle it cannot be determined unless we had 2 angles, since it can have highly acute angles.

7 0
4 years ago
Read 2 more answers
SOLVE THE SYSTEM OF EQUATIONS <br> −2x+6y=−38 <br> 3x−4y=32
WARRIOR [948]

X= 4 and Y =-5 im not really 100% sure but its all i know

6 0
4 years ago
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A Ferris wheel at the local fair has a diameter of 36 feet and a midline of 15 feet. This Ferris wheel makes one revolution ever
AVprozaik [17]

Answer:

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5 0
3 years ago
(2pm^-1q^0)^-4 • 2m ^-1 p^3 / 2pq^2
Montano1993 [528]

Answer:

\dfrac{m^3}{16p^2q^2}

Step-by-step explanation:

Given:

(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}

1.

m^{-1}=\dfrac{1}{m}

2.

q^0=1

3.

2pm^{-1}q^0=2p\cdot \dfrac{1}{m}\cdot 1=\dfrac{2p}{m}

4.

(2pm^{-1}q^0)^{-4}=\left(\dfrac{2p}{m}\right)^{-4}=\left(\dfrac{m}{2p}\right)^4=\dfrac{m^4}{(2p)^4}=\dfrac{m^4}{16p^4}

5.

m^{-1}=\dfrac{1}{m}

6.

2m^{-1} p^3=2\cdot \dfrac{1}{m}\cdot p^3=\dfrac{2p^3}{m}

7.

\dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{\frac{2p^3}{m}}{2pq^2}=\dfrac{2p^3}{m}\cdot \dfrac{1}{2pq^2}=\dfrac{p^2}{mq^2}

8.

(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{m^4}{16p^4}\cdot \dfrac{p^2}{mq^2}=\dfrac{m^3}{16p^2q^2}

8 0
3 years ago
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