I need quick! 20 points!

1 answer:

7 0

Given is h(x) = 4x − 5 and t(x) = 2x + 4.

Part A: Finding (h + t)(x)
$(h + t)(x) = (4x - 5) + (2x + 4) \\ (h + t)(x) = 4x - 5 + 2x + 4 \\ (h + t)(x) = 4x + 2x - 5 + 4 \\ (h + t)(x) = 6x - 1$

Part B: Finding (h ⋅ t)(x)
$(h ⋅ t)(x) = (4x - 5)(2x + 4) \\ (h ⋅ t)(x) = 4x(2x + 4) - 5(2x + 4) \\ (h ⋅ t)(x) = 8 {x}^{2} + 16x - 10x - 20 \\ (h ⋅ t)(x) = 8 {x}^{2} + 6x - 20$
Part C: Finding h[t(x)]
$h[t(x)] = 4 \times t(x) - 5 \\ h[t(x)] = 4(2x + 4) - 5 \\ h[t(x)] = 8x + 16 - 5 \\ h[t(x)] = 8x + 11$

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Step-by-step explanation:

When you know the length of the three sides of the triangle and not a single angle, the law of sines is of no use. You need to use the law of cosines. In your case, since you want to find angle "C", you use the following;

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