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Ber [7]
3 years ago
7

-b - 7?

Mathematics
1 answer:
Olegator [25]3 years ago
8 0

Answer:

4 better xplanation inbox

Step-by-step explanation:

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Helpp me plssssss????
snow_tiger [21]

Answer:

C

Step-by-step explanation:

The radius is 4 making the diameter 16 you need the other side since in half so that leaves you with answer C...16\pi yd^2

3 0
2 years ago
Write the answer in scientific notation and in standard form.Round to the appropriate number of significant digits.
Aleks [24]

Hey there, the answer is C.

After multiplying, you will get .37278. However, it's asking you to consider the significant figure of 3 so you'll have to round it to .373. C has that only option

7 0
3 years ago
Determine the first five multiples for the following number 22
iragen [17]

Lets find

\\ \sf\longmapsto M_n=22n

\\ \sf\longmapsto M_1=22(1)=22

\\ \sf\longmapsto M_2=22(2)=44

\\ \sf\longmapsto M_3=22(3)=66

\\ \sf\longmapsto M_4=22(4)

\\ \sf\longmapsto M_4=88

\\ \sf\longmapsto M_5=22(5)=110

3 0
3 years ago
L
seraphim [82]
A parallel line has the same gradient

First you will need to rearrange the equation 10x+2y=-2

Then once you do that please comment

If you don’t know how please comment
7 0
2 years ago
Read 2 more answers
1 + tanx / 1 + cotx =2
Lera25 [3.4K]

Answer:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or x = tan^(-1)(-(i sqrt(3))/2 + 1/2) + π n_2 for n_2 element Z

Step-by-step explanation:

Solve for x:

1 + cot(x) + tan(x) = 2

Multiply both sides of 1 + cot(x) + tan(x) = 2 by tan(x):

1 + tan(x) + tan^2(x) = 2 tan(x)

Subtract 2 tan(x) from both sides:

1 - tan(x) + tan^2(x) = 0

Subtract 1 from both sides:

tan^2(x) - tan(x) = -1

Add 1/4 to both sides:

1/4 - tan(x) + tan^2(x) = -3/4

Write the left hand side as a square:

(tan(x) - 1/2)^2 = -3/4

Take the square root of both sides:

tan(x) - 1/2 = (i sqrt(3))/2 or tan(x) - 1/2 = -(i sqrt(3))/2

Add 1/2 to both sides:

tan(x) = 1/2 + (i sqrt(3))/2 or tan(x) - 1/2 = -(i sqrt(3))/2

Take the inverse tangent of both sides:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or tan(x) - 1/2 = -(i sqrt(3))/2

Add 1/2 to both sides:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or tan(x) = 1/2 - (i sqrt(3))/2

Take the inverse tangent of both sides:

Answer:  x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or x = tan^(-1)(-(i sqrt(3))/2 + 1/2) + π n_2 for n_2 element Z

4 0
3 years ago
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