Help Pleaseeee. Find the hypotenuse.

I just need no. a please help me to prove this.

OleMash [197]

Answer: see proof below

<u>Step-by-step explanation:</u>

Given: A + B + C = π and cos A = cos B · cos C

scratchwork:

A + B + C = π

A = π - (B + C)

cos A = cos [π - (B + C)] Apply cos

= - cos (B + C) Simplify

= -(cos B · cos C - sin B · sin C) Sum Identity

= sin B · sin C - cos B · cos C Simplify

cos B · cos C = sin B · sin C - cos B · cos C Substitution

2cos B · cos C = sin B · sin C Addition

$2=\dfrac{\sin B\cdot \sin C}{\cos B \cdot \cos C}$ Division

2 = tan B · tan C

$\text{Use the Sum Identity:}\quad \tan(B+C)=\dfrac{\tan B+\tan C}{1-\tan B\cdot \tan C}$

<u>Proof LHS → RHS</u>

Given: A + B + C = π

Subtraction: A = π - (B + C)

Apply tan: tan A = tan(π - (B + C))

Simplify: = - tan (B + C)

$\text{Sum Identity:}\qquad \qquad \qquad =-\bigg(\dfrac{\tan B+\tan C}{1-\tan B\cdot \tan C}\bigg)$

Substitution: = -(tan B + tan C)/(1 - 2)

Simplify: = -(tan B + tan C)/-1

= tan B + tan C

LHS = RHS: tan B + tan C = tan B + tan C $\checkmark$

5 0

2 years ago

ASAP!!! PLEASE HELP & EXPLAIN.

Vika [28.1K]

Answer:

B

Step-by-step explanation:

the domain of a graph consists of all the input values shown on the x-axis.

so because the only parts shown on the x axis are 6 and -6 that is the answer!

hope this helps!

4 0

1 year ago

Can anybody help pls

Gekata [30.6K]

Answer:

2t

Step-by-step explanation:

The expressions are not the same because different terms are negative. We can visualize the difference by rewriting the expressions as 13t + (-2t) and (-13t) + 2t. In the first expression, 13t-2t, the 2t is negative because it is being subtracted, and subtracting a number is the same as adding a negative.

6 0

2 years ago

Read 2 more answers

Which of the following methods would be the easiest to use to solve x2 + 4x + 3 = 0?

tresset_1 [31]

It would be easier to do A

6 0

3 years ago

Read 2 more answers

Q‒1. [5×4 marks] a) How many three-digit numbers can be formed from the digits 0, 1, 2, 3, 4, 5, and 6? (150) b) How many three-

amid [387]

Answer:

a) 294

b) 180

c) 75

d) 174

e) 105

Step-by-step explanation:

I assume that for each problem, the first digit can't be 0.

a) There are 6 digits that can be first, 7 digits that can be second, and 7 digits that can be third.

6×7×7 = 294

b) This time, no digit can be used twice, so there are 6 digits that can be first, 6 digits that can be second, and 5 digits that can be third.

6×6×5 = 180

c) Again, each digit can only be used once, but this time, the last digit must be odd.

If only the last digit is odd, there are 3×3×3 = 27 possible numbers.

If the first and last digits are odd, there are 3×4×2 = 24 possible numbers.

If the second and last digits are odd, there are 3×3×2 = 18 possible numbers.

If all three digits are odd, there are 3×2×1 = 6 possible numbers.

The total is 27 + 24 + 18 + 6 = 75.

d) If the first digit is 3, and the second digit is 3, there are 1×1×6 = 6 possible numbers.

If the first digit is 3, and the second digit is greater than 3, there are 1×3×7 = 21 possible numbers.

If the first digit is greater than 3, there are 3×7×7 = 147 numbers.

The total is 6 + 21 + 147 = 174.

e) If the first digit is 3, and the second digit is greater than 3, then there are 1×3×5 = 15 possible numbers.

If the second digit is greater than 3, there are 3×6×5 = 90 possible numbers.

The total is 15 + 90 = 105.

6 0

3 years ago