Given the triangles ABC and PQR.
Angle A = Angle P
Angle B = Angle Q
Angle C = Angle R

Angle B = 3v+4
Angle Q = 8v-6
Let's find v

3v+4 = 8v-6
-5v = -10
v = 2

Angle B = 3v + 4
Angle B = 3(2) + 4
Angle B = 10.

The correct answer is letter C. 10

8 0

3 years ago

PLEASE HELP!!! TRIG, Theres a picture!!

BartSMP [9]

Answer:

<h2>The answer is option C</h2>

Step-by-step explanation:

$3( { \tan }^{2} \theta - { \sec }^{2} \theta)$

Using trigonometric identities

That's

$1 + { \tan }^{2} \theta = { \sec }^{2} \theta$

Rewrite the expression

That's

$3({ \tan }^{2} \theta - (1 + { \tan }^{2} \theta)) \\$

Simplify

We have

$3({ \tan }^{2} \theta - 1 - { \tan }^{2} \theta) \\ 3({ \tan }^{2} \theta - { \tan }^{2} \theta - 1)$

So we have

3( - 1)

We have the final answer as

Hope this helps you

3 0

3 years ago

Factor:<br> 4x^2+4xy−288y^2

Sonja [21]

$4x^2+4xy-288y^2=\\
4(x^2+xy-72y)=\\
4(x^2+9xy-8xy-72y)=\\
4(x(x+9y)-8y(x+9y))=\\
4(x-8y)(x+9y)$

7 0

3 years ago

Solve.<br> - 5 - 3 1 / 2 = 0<br> 21<br> Done

patriot [66]

This equation does not have an answer

5 0

3 years ago