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inessss [21]
3 years ago
9

Solve for each indicated angle measure or variable in the figure m x

Mathematics
1 answer:
ruslelena [56]3 years ago
4 0
84° and 4x are alternative angles. so 4×+84= 180 since a straight line is 180° solve the equation to get x=24. do the same with m. 4 (24) +m=180 so 96+m=180. m=84°. sorry if this is wrong. but im 99% sure its correct.
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djverab [1.8K]

problem decoded dude

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6 0
3 years ago
Read 2 more answers
Write the slope intercept form that satisfies the following conditions
Olin [163]

y = 3x - 13

Step-by-step explanation:

The line will have the same slope as the given line, which is m = 3. Therefore, the equation of the line in its slope-intercept form is

y = 3x + b

Next we need to solve for b by plugging in the coordinates of the given point:

(-4) = 3(3) + b \Rightarrow b = -13

So the equation of the line is

y = 3x - 13

7 0
3 years ago
A baseball player hit 65 home runs in a season. Of the 65 home​ runs, 19 went to right​ field, 23 went to right-center​ field, 9
olchik [2.2K]

Answer:

a) 29.23% probability that a randomly selected home run was hit to right field

b) 29.23% probability that a randomly selected home run was hit to right field, which is not lower than 5% nor it is higher than 95%. So it was not unusual for this player to hit a home run to right field.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes. It is said to be unusual if it is lower than 5% or higher than 95%.

(a) What is the probability that a randomly selected home run was hit to right field?

Desired outcomes:

19 home runs hit to right field

Total outcomes:

65 home runs

19/65 = 0.2923

29.23% probability that a randomly selected home run was hit to right field

(b) Was it unusual for this player to hit a home run to right field?

29.23% probability that a randomly selected home run was hit to right field, which is not lower than 5% nor it is higher than 95%. So it was not unusual for this player to hit a home run to right field.

8 0
4 years ago
Express the fifth roots of unity in standard form a + bi. with 1 + 0i
marishachu [46]

Answer:

For K=0

cos(\frac{0+2\pi 0}{5})+isin(\frac{0+2\pi 0}{5})\\cos(0)+isin(0)=1+i0

For K=1:

cos(\frac{0+2\pi 1}{5})+isin(\frac{0+2\pi 1}{5})\\cos(2\pi/5)+isin(2\pi/5)=0.3090+i0.9510

For K=2:

cos(\frac{0+2\pi 2}{5} )+isin(\frac{0+2\pi 2}{5} )\\cos(4\pi/5)+isin(4\pi/5)=-0.809+i0.587

For K=3:

cos(\frac{0+2\pi 3}{5} )+isin(\frac{0+2\pi 3}{5} )\\cos(6\pi/5)+isin(6\pi/5)=-0.809-i0.587

For K=4:

cos(\frac{0+2\pi 4}{5} )+isin(\frac{0+2\pi 4}{5} )\\cos(8\pi/5)+isin(8\pi/5)=0.3091-i0.9510

Step-by-step explanation:

Fifth Root is given by:

\sqrt[5]{z}=1+0i

The above equation will become:

z=(1+0i)^5

It can be written as:

z=[cos(0)+isin(0)]^5

|z|=1,

According to De-moivre's Theorem:

z=cos(\frac{0}{5})+isin(\frac{0}{5})\\  z=cos(0)+isin(0)

Now, Fifth Roots of unity in standard form a + bi :

\sqrt[5]{z}=[{cos(0+2\pi k)+isin(0+2\pi k)}]^{1/5}

k=0,1,2,3,4

For K=0

cos(\frac{0+2\pi 0}{5})+isin(\frac{0+2\pi 0}{5})\\cos(0)+isin(0)=1+i0

For K=1:

cos(\frac{0+2\pi 1}{5})+isin(\frac{0+2\pi 1}{5})\\cos(2\pi/5)+isin(2\pi/5)=0.3090+i0.9510

For K=2:

cos(\frac{0+2\pi 2}{5} )+isin(\frac{0+2\pi 2}{5} )\\cos(4\pi/5)+isin(4\pi/5)=-0.809+i0.587

For K=3:

cos(\frac{0+2\pi 3}{5} )+isin(\frac{0+2\pi 3}{5} )\\cos(6\pi/5)+isin(6\pi/5)=-0.809-i0.587

For K=4:

cos(\frac{0+2\pi 4}{5} )+isin(\frac{0+2\pi 4}{5} )\\cos(8\pi/5)+isin(8\pi/5)=0.3091-i0.9510

6 0
3 years ago
*2.6-(5.85)<br> Evaluate It
daser333 [38]

Answer:

-3.25

Step-by-step explanation:

4 0
4 years ago
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