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alexira [117]
3 years ago
8

Find m3 if m5 = 38° and m6 = 62°.

Mathematics
2 answers:
geniusboy [140]3 years ago
5 0
M5 minus M6 will get M1
So 62 - 38 = 24
24 x 3 = 72
So M3 is 72
Levart [38]3 years ago
3 0

the answer is :

m3 = 80°

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Make g the subject of this relation​
zhannawk [14.2K]

Answer:

see explanation

Step-by-step explanation:

Simplify the radical

\frac{U}{\frac{1}{f}+\frac{1}{g}  }

= \frac{U}{\frac{g+f}{fg} }

= \frac{Ufg}{g+f}

Square both sides

T² = \frac{Ufg}{g+f} ( multiply both sides by (g + f) )

T²(g + f) = Ufg ( distribute left side )

T²g + T²f = Ufg ← subtract Ufg from both sides

T²g - Ufg + T²f = 0 ← subtract T²f from both sides

T²g - Ufg = - T²f ← factor out g from each term on the left side

g(T² - Uf) = - T²f ← divide both sides by (T² - Uf)

g = - \frac{T^2f}{T^2-Uf} = \frac{T^2f}{Uf-T^2}

6 0
3 years ago
What is (a+2)(a+2) make sure to multiply
a_sh-v [17]

Answer:

a² + 4 a + 4

Step-by-step explanation:

( a + 2 ) ( a + 2 )

Expand brackets

a² + 2 a + 2 a + 4

Simplify

a² + 4 a + 4

7 0
2 years ago
Read 2 more answers
Find the midpoint of (1, 3) and (6, 5)
olasank [31]

Answer:

(3.5, 4)

Step-by-step explanation:

add and then divide (x1+x2 / 2, y1 + y2 / 2) so [(1+6)/2, (3+5)/2], so (3.5, 4)

5 0
3 years ago
Suppose we roll a fair six-sided die and sum the values obtained on each roll, stopping once our sum exceeds 307. Approximate th
SOVA2 [1]

Answer:

There is a probability of P=0.94 that at least 81 rolls are needed to get the sum of 307.

Step-by-step explanation:

The roll of a six-sided die sum has this parameters:

Mean: 3.5

Standard deviation: 1.7

If the die is rolled 81 times, the distribution of the sum of the 81 rolls will have the following parameters:

\mu=n*M=81*3.5=283.5\\\\\sigma=\sqrt{n}s=\sqrt{81}*1.7=9*1.7=15.3

Note: the variance of the sum of random variables is equal to the sum of the variance of the individual variables.

Then, we can calculate the probabilties that the sum of 81 rolls is lower than 307.

z=\frac{x-\mu}{\sigma}=\frac{307-283.5}{15.3}=  \frac{23.5}{15.3}= 1.536\\\\\\P(x

There is 94% of chances that the sum of 81 rolls is lower than 307, so there is a probability of P=0.94 that at least 81 rolls are needed to get the sum of 307.

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3 years ago
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Step-by-step explanation:

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