Answer:
Refer to attached
Step-by-step explanation:
<u>Given inequality</u>
<u>We can put this as</u>
We know the given exponential function has positive range y > 0
And the linear inequality covers the space above the line
This is all matching the last graph that is attached
Answer:
Becky completed the proof incorrectly.
Step-by-step explanation:
The only difference of the two proofs, at least what I can see, is the reason for step 2. Angie claims that it is the Definition of Supplementary Angles, while Becky claims Angle Addition Postulate.
The Definition of Supplementary Angles states that m∠A + m∠B = 180°.
While the Angle Addition Postulate states that m∠A + m∠B = m∠C.
Because the context used is that m∠AKG + m∠GKB = 180° and m∠GKB + m∠HKB = 180°, it would be using the Definition of Supplementary Angles.
So Becky is the one who is incorrect.
(5^-2 ● 4^-4)^-2
you need to multiply tge power of tge outside the bracket with tge power inside the bracket to eliminate the power
= 5^4 ● 4^8
the answer is A
Answer: see proof below
<u>Step-by-step explanation:</u>
Given: A + B + C = 90° → A + B = 90° - C
→ C = 90° - (A + B)
Use the Double Angle Identity: cos 2A = 1 - 2 sin² A
→ sin² A = (1 - cos 2A)/2
Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · cos [(A - B)/2]
Use the Product to Sum Identity: cos (A - B) - cos (A + B) = 2 sin A · sin B
Use the Cofunction Identities: cos (90° - A) = sin A
sin (90° - A) = cos A
<u>Proof LHS → RHS:</u>
LHS: sin² A + sin² B + sin² C

![\text{Sum to Product:}\quad 1-\dfrac{1}{2}\bigg[2\cos \bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A-2B}{2}\bigg)\bigg]+\sin^2 C\\\\\\.\qquad \qquad \qquad =1-\cos (A+B)\cdot \cos (A-B)+\sin^2 C](https://tex.z-dn.net/?f=%5Ctext%7BSum%20to%20Product%3A%7D%5Cquad%201-%5Cdfrac%7B1%7D%7B2%7D%5Cbigg%5B2%5Ccos%20%5Cbigg%28%5Cdfrac%7B2A%2B2B%7D%7B2%7D%5Cbigg%29%5Ccdot%20%5Ccos%20%5Cbigg%28%5Cdfrac%7B2A-2B%7D%7B2%7D%5Cbigg%29%5Cbigg%5D%2B%5Csin%5E2%20C%5C%5C%5C%5C%5C%5C.%5Cqquad%20%5Cqquad%20%5Cqquad%20%3D1-%5Ccos%20%28A%2BB%29%5Ccdot%20%5Ccos%20%28A-B%29%2B%5Csin%5E2%20C)
Given: 1 - cos (90° - C) · cos (A - B) + sin² C
Cofunction: 1 - sin C · cos (A - B) + sin² C
Factor: 1 - sin C [cos (A - B) + sin C]
Given: 1 - sin C[cos (A - B) - sin (90° - (A + B))]
Cofunction: 1 - sin C[cos (A - B) - cos (A + B)]
Sum to Product: 1 - sin C [2 sin A · sin B]
= 1 - 2 sin A · sin B · sin C
LHS = RHS: 1 - 2 sin A · sin B · sin C = 1 - 2 sin A · sin B · sin C 