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viktelen [127]
3 years ago
14

I really need help with figuring out how to calculate b, I've been really confused.

Mathematics
1 answer:
aleksley [76]3 years ago
3 0

Answer:

The value of b is -12

Step-by-step explanation:

Differentiate the given function with respect to x.

f'(x)=4x-4

Subsitute 3 for x in f'(x).

f'(3)=4\cdot3-4\\=12-4\\=8

The slope of a tangent line is f'(3)=m=8.

Subsitute 8 for m, 12 for y_1 and 3 for x_1 in y-y_1=m(x-x_1).

y-12=8(x-3)\\y-12=8x-24\\y=8x-24+12\\y=8x-12

So, the value of b is -12

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

Given: cos 330 = \frac{\sqrt3}{2}

Use the Double-Angle Identity: cos 2A = 2 cos² A - 1

\text{Scratchwork:}\quad \bigg(\dfrac{\sqrt3 + 2}{2\sqrt2}\bigg)^2 = \dfrac{2\sqrt3 + 4}{8}

Proof LHS → RHS:

LHS                          cos 165

Double-Angle:        cos (2 · 165) = 2 cos² 165 - 1

                             ⇒ cos 330 = 2 cos² 165 - 1

                             ⇒ 2 cos² 165  = cos 330 + 1

Given:                        2 \cos^2 165  = \dfrac{\sqrt3}{2} + 1

                              \rightarrow 2 \cos^2 165  = \dfrac{\sqrt3}{2} + \dfrac{2}{2}

Divide by 2:               \cos^2 165  = \dfrac{\sqrt3+2}{4}

                             \rightarrow \cos^2 165  = \bigg(\dfrac{2}{2}\bigg)\dfrac{\sqrt3+2}{4}

                             \rightarrow \cos^2 165  = \dfrac{2\sqrt3+4}{8}

Square root:             \sqrt{\cos^2 165}  = \sqrt{\dfrac{4+2\sqrt3}{8}}

Scratchwork:            \cos^2 165  = \bigg(\dfrac{\sqrt3+1}{2\sqrt2}\bigg)^2

                             \rightarrow \cos 165  = \pm \dfrac{\sqrt3+1}{2\sqrt2}

             Since cos 165 is in the 2nd Quadrant, the sign is NEGATIVE

                             \rightarrow \cos 165  = - \dfrac{\sqrt3+1}{2\sqrt2}

LHS = RHS \checkmark

4 0
4 years ago
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