Problem One
Sin(90 + q) = sin(90)*cos(q) + sin(q)*cos(90)
sin(90 + q) = 1 * cos(q) + sin(q)*0
sin(90 + q) = cos(q)

You can elimate anything beginning with Cos
You can also eliminate the sin function with a minus between it.

Problem 2
Sin(x) = 1/csc(x) That makes C and D incorrect.
Correct answer has to be A. Since this is a multiple choice question you can't qualify the answer, but there are some points that exhibit questionable behavior, like 2 $\pi$ *f*t = 0 for example. The function becomes undefined.

4 0

3 years ago

What would the markup be???

Alja [10]

Hello!

To find the markup you have to subtract the selling price from the store price so

91 - 65 = 26

So there is a $26 markup

Hope this helps!

5 0

3 years ago

Come help a sister out on 2 geometry questions please

PSYCHO15rus [73]

1- question

Data:
width = 6 cm
length = 5 cm
height = 8 cm

First we will find the area of the base that is square, we have:

Ab = 6 * 5
Ab = 30 cm²

Now we will calculate the volume of a square pyramid, we have:
Formula: $V = \frac{1}{3} A_{b} *h$

Solving:
$V = \frac{1}{3} A_{b} *h$
$V = \frac{1}{3}*30*8$
$V = \frac{240}{3}$
$\boxed{\boxed{V = 80\:cm^3}}\end{array}}\qquad\quad\checkmark$

Answer:
D) 80 cm³

2- question:

Data:
width = 7 in
length = 7 in
height = 9 in

First we will find the area of the base that is square, we have:

Ab = 7 * 7
Ab = 49 in²

Now we will calculate the volume of a square pyramid, we have:
Formula: $V = \frac{1}{3} A_{b} *h$

Solving:
$V = \frac{1}{3} A_{b} *h$
$V = \frac{1}{3}*49*9$
$V = \frac{441}{3}$
$\boxed{\boxed{V = 147\:in^3}}\end{array}}\qquad\quad\checkmark$

Answer:
C) 147 in³

3 0

3 years ago