2 years ago

Help with exam 11 please

1 answer:

Feliz [49]2 years ago

5 0

Answer:

Step-by-step explanation:

Hope this helps u!!

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Serga [27]

I’m not sure on the answer choices but this is what I got when I did synthetic division.

8 0

3 years ago

Ainat [17]

Recall the double angle identity for cosine:

$\cos(2x) = \cos^2(x) - \sin^2(x) = 1 - 2 \sin^2(x)$

It follows that

$\sin^2(x) = \dfrac{1 - \cos(2x)}2 \implies \sin(x) = \pm \sqrt{\dfrac{1-\cos(2x)}2} \implies \csc(x) = \pm \sqrt{\dfrac2{1-\cos(2x)}}$

Since 0° < 22° < 90°, we know that sin(22°) must be positive, so csc(22°) is also positive. Let x = 22°; then the closest answer would be C,

$\csc(22^\circ) = \sqrt{\dfrac2{1-\cos(44^\circ)}} = \sqrt{\dfrac2{1-\frac5{13}}} = \dfrac{\sqrt{13}}2$

but the problem is that none of these claims are true; cot(32°) ≠ 4/3, cos(44°) ≠ 5/13, and csc(22°) ≠ √13/2...

3 0

2 years ago

Annette [7]

False hope that helps

3 0

2 years ago

Ede4ka [16]

Answer:

it a

Step-by-step explanation:

4 0

3 years ago

zaharov [31]

Answer:

k+7

Step-by-step explanation:

because k is the age of kit so Anika is 7 years older, so u have to add

sorry if I'm wrong

7 0

3 years ago