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svetoff [14.1K]
3 years ago
5

How many parts are needed to prove triangles are similar?

Mathematics
1 answer:
Alex73 [517]3 years ago
6 0
If you have two sides, you only need two. Otherwise it’s 3 parts
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Find WV 2X+26 2X+31
blsea [12.9K]

Answer:

4

Step-by-step explanation:

Yes u are doing it right

So just find x first

2x + 26 + 2x +31 =13

4x + 57 = 13

4x = - 44

x = - 11

Since now u know x = - 11, and u know

WV =2X+26

So...

2x + 26

=2(-11) + 26

= - 22 + 26

= 4

3 0
3 years ago
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
Question is in the photo, please no links we know they are viruses.
prisoha [69]

Answer:

A

Step-by-step explanation:

1) Multiply 0.75 for m and -250

162.5 + 0.75m - 187.5

2) sum the therms without m

0.75m - 25

6 0
3 years ago
What <br> is (81m^6)^1/2 simplified
ANEK [815]
(81m^6)^1/2 = (3^4*1/2) (m^6*1/2) = 3²m³
7 0
3 years ago
What is the radius of a circle with a sector area of 7pi sq. ft. and an arc whose measure is 70°? 6 feet 10 feet 18 feet
Svetlanka [38]

Here we have to use the formula of area of sector, which is

A = (\frac{\Theta}{360})*\pi r^2

Theta = 70 degree, and radius is unknown here. It is given that area = 7 pi .

Substituting the given values in the formula, we will get

7 \pi = (\frac{70}{360})*\pi r^2

\frac{7 \pi * 360}{70 \pi}=r^2

36 = r^2=>r=6ft

Therefore the correct option is the first option .

5 0
3 years ago
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