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3 years ago

How do you solve for #23

Mathematics

1 answer:

Bingel [31]3 years ago

8 0

5 is you answer because absolute value is the distance on a number line from 0

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Can i get the answer to this question please???? and fast

lord [1]

The triangle must be isosceles, due to the perpendicular bisector. Therefore, the expression 20 - 15 can be used to yield an answer of B. 5.

Hope this helps!

5 0

3 years ago

HELP ASAP

Ilya [14]

Answer:

Step-by-step explanation:

it is an adjacent/linear pair

5 0

3 years ago

Simplify 3/4 - 1 1/4

lbvjy [14]

I think the answer is -7/4.

4 0

3 years ago

Read 2 more answers

Using the Distributive Property to factorize the equation 3x^2 + 24x = 0, you get _____The solution of the equation is _____

stiv31 [10]

Solution is x=0 x=-8

7 0

3 years ago

Please prove this........

Crazy boy [7]

Answer: see proof below

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

Given: A + B + C = π → C = π - (A + B)

→ sin C = sin(π - (A + B)) cos C = sin(π - (A + B))

→ sin C = sin (A + B) cos C = - cos(A + B)

Use the following Sum to Product Identity:

sin A + sin B = 2 cos[(A + B)/2] · sin [(A - B)/2]

cos A + cos B = 2 cos[(A + B)/2] · cos [(A - B)/2]

Use the following Double Angle Identity:

sin 2A = 2 sin A · cos A

<u>Proof LHS → RHS</u>

LHS: (sin 2A + sin 2B) + sin 2C

$\text{Sum to Product:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-\sin 2C$

$\text{Double Angle:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-2\sin C\cdot \cos C$

$\text{Simplify:}\qquad \qquad 2\sin (A + B)\cdot \cos (A - B)-2\sin C\cdot \cos C$

$\text{Given:}\qquad \qquad \quad 2\sin C\cdot \cos (A - B)+2\sin C\cdot \cos (A+B)$

$\text{Factor:}\qquad \qquad \qquad 2\sin C\cdot [\cos (A-B)+\cos (A+B)]$

$\text{Sum to Product:}\qquad 2\sin C\cdot 2\cos A\cdot \cos B$

$\text{Simplify:}\qquad \qquad 4\cos A\cdot \cos B \cdot \sin C$

LHS = RHS: 4 cos A · cos B · sin C = 4 cos A · cos B · sin C $\checkmark$

7 0

3 years ago