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3 years ago

9

Find the sin P rounded to the nearest hundredth

1 answer:

Lady_Fox [76]3 years ago

3 0

Answer:

SOH-CAH-TOA

$\sin \left(x\right)=\frac{6}{\sqrt{49+36}}$ = 40.60°

SOH = SIN = OPP/HYP

SIN(Θ) = 6/ $\sqrt{49+36 }$

Step-by-step explanation:

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At midnight, the temperature was 30 degrees Fahrenheit. By 6:00 am, IT had dropped 5 degrees Fahrenheit and by noon, it had incr

Allisa [31]

At noon, the temperature was 36 degrees. At midnight, the temperature was 30 degrees. By 6:00 am it drooped 5 degrees. Since it dropped 5 degrees, you subtract 5 from 30, which equals 25. It then increased by 11 degrees. Since the temperature increased, you add 11 to 25, which equals 36. At noon, the temperature was 36 degrees.

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A universal set U consists of elements. If sets A, B, and C are proper subsets of U and n(U), n(A B)n(A C)n(B C), n(A

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Answer:

$n(A\ u\ B) = 10$

$n(A'\ u\ C) = 10$

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Step-by-step explanation:

Given

$n(U) = 12$

$n(A\ n\ B) =n(A\ n\ C) = n(B\ n\ C) =6$

$n(A\ n\ B\ n\ C)=4$

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Required

Solve a, b and c

There are several ways to solve this; the best is by using Venn diagram (see attachment for diagram)

Solving (a):

$n(A\ u\ B)$

This is calculated as:

$n(A\ u\ B) = n(A) + n(B) - n(A\ n\ B)$

From the attachment

$n(A) = 0+2+4+2 = 8$

$n(B) = 0+2+4+2 = 8$

$n(A\ n\ B) = 4 +2 = 6$

So:

$n(A\ u\ B) = 8 + 8 - 6$

$n(A\ u\ B) = 10$

Solving (b):

$n(A'\ u\ C)$

This is calculated as:

$n(A'\ u\ C) = n(A') + n(C) -n(A'\ n\ C)$

From the attachment

$n(A) = n(U) - n(A) = 12 - 8 = 4$

$n(C) = 0+2+4+2 = 8$

$n(A'\ n\ C) = 2$

So:

$n(A'\ u\ C) = 4 + 8 - 2$

$n(A'\ u\ C) = 10$

Solving (c):

$n(A\ n\ B)'$

This is calculated as:

$n(A\ n\ B)' = n(U) - n(A\ n\ B)$

$n(A\ n\ B)' = 12- 6$

$n(A\ n\ B)' = 6$

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Answer:

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Step-by-step explanation:

2.18 × 10^4

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