Now all you have to do is figure out the range which is the y value. At first glance I think it might be 3, but that does not look very logical. I'll post this much of it now and be back in under an hour with a more complete answer.

Of course! How silly of me. There is a minimum of y = 1 in the range which comes from x = 0

I've included a graph so you can see how this all works.

So the range = 1 ≤ y ≤ 3

5 0

2 years ago

How do you solve for "r" in the literal equation "I=Prt"

Mrac [35]

R=I/Pt
hopefully this helps

8 0

2 years ago

Please someone help me to prove this.

morpeh [17]

Answer: see proof below

Step-by-step explanation:

Use the Double Angle Identity: sin 2Ф = 2sinФ · cosФ

Use the Sum/Difference Identities:

sin(α + β) = sinα · cosβ + cosα · sinβ

cos(α - β) = cosα · cosβ + sinα · sinβ

Use the Unit circle to evaluate: sin45 = cos45 = √2/2

Use the Double Angle Identities: sin2Ф = 2sinФ · cosФ

Use the Pythagorean Identity: cos²Ф + sin²Ф = 1

Proof LHS → RHS

LHS: 2sin(45 + 2A) · cos(45 - 2A)

Sum/Difference: 2 (sin45·cos2A + cos45·sin2A) (cos45·cos2A + sin45·sin2A)

Unit Circle: 2[(√2/2)cos2A + (√2/2)sin2A][(√2/2)cos2A +(√2/2)·sin2A)]

Expand: 2[(1/2)cos²2A + cos2A·sin2A + (1/2)sin²2A]

Distribute: cos²2A + 2cos2A·sin2A + sin²2A

Pythagorean Identity: 1 + 2cos2A·sin2A

Double Angle: 1 + sin4A

LHS = RHS: 1 + sin4A = 1 + sin4A $\checkmark$

6 0

2 years ago