Domain and range of 10x - 2y = 4

Mathematics

1 answer:

Anna71 [15]3 years ago

5 0

Answer:

domain: (-∞,∞)

range: (-∞,∞)

Step-by-step explanation:

You need to convert this into a linear equation (y=mx+b)

So we add 2y and divide by 2 to get y by itself

that gives us 5x-2=y

If you pictures this graph in your head you'll see that the domain and range are all real numbers

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Please help me to prove this! I need is no.(c). So, please help me do it.

zloy xaker [14]

Answer: see proof below

Step-by-step explanation:

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

Proof LHS → RHS:

LHS: sin² A + sin² B + sin² C

$\text{Double Angle:}\qquad \dfrac{1-\cos 2A}{2}+\dfrac{1-\cos 2B}{2}+\sin^2 C\\\\\\.\qquad \qquad \qquad =\dfrac{1}{2}\bigg(2-\cos 2A-\cos 2B\bigg)+\sin^2 C\\\\\\.\qquad \qquad \qquad =1-\dfrac{1}{2}\bigg(\cos 2A+\cos 2B\bigg)+\sin^2 C$

$\text{Sum to Product:}\quad 1-\dfrac{1}{2}\bigg[2\cos \bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A-2B}{2}\bigg)\bigg]+\sin^2 C\\\\\\.\qquad \qquad \qquad =1-\cos (A+B)\cdot \cos (A-B)+\sin^2 C$