All categories

3 years ago

Sin(42) = cos(x) Solve for x

Mathematics

1 answer:

mariarad [96]3 years ago

3 0

Answer:

X = 48

Step-by-step explanation:

When you type sin(42) in your calculator it will give you something around 0.669.

If you try reverse sinus sin-1(0.669) it will give you 42.

Therefore you can say that:

Cos(x) = 0.669

So you can reverse the cos to get your answer

Cos-1(0.669) = 48

Cos-1 (sin(42))

You might be interested in

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Nady [450]

Answer:

A) 2x³+11x²+8x-16

Step-by-step explanation:

When you multiply s(x) by t(x) you get something like this:

$s(x) \times t(x) = (2 {x}^{2} + 3x - 4) \times (x + 4) \\ = 2 {x}^{3} + 3 {x}^{2} - 4x + 8 {x}^{2} + 12x - 16 \\ = 2 {x}^{3} + 11 {x}^{2} + 8x - 16$

5 0

3 years ago

Read 2 more answers

A quadrilateral has vertices a(-4 -8) b (-2, 34) c(6,12 and d (10, -30). It undergoes the transformation (x,y) to (x/2,y/2).

worty [1.4K]

Answer:

Step-by-step explanation:

scale factor:

1/2

the coordinates of the new vertices are

A (-4/2, -8/2)

A:(-2, -4)

B (-2/2, 34/2)

B: (-1, 17)

C (6/2, 12/2)

C:(3, 6)

D (10/2, -30/2)

D:(5, -15)

I guess the dilation is 2 times bigger?

4 0

3 years ago

Which of the following are the coordinates of vertices of the following square centered at the Origin, with side length b?

adoni [48]

Answer:

Option (3) is correct.

coordinate of given square centered at the Origin, with side length b is W $(\frac{b}{2},\frac{b}{2})$ , S $(\frac{-b}{2},\frac{b}{2})$ , T $(\frac{-b}{2},\frac{-b}{2})$ and Z $(\frac{b}{2},\frac{b}{2})$ .

Step-by-step explanation:

Given a square centered at the Origin, with side length b.

We have to find the coordinates of vertices of the square.

Since, the given square is centered at origin (0,0) and length of side is b then x and y axis divide each side in equal part.

Thus, each coordinate of square is $\frac{b}{2}$ .

Since, we know in first quadrant both x and y are positive thus, point W has coordinate $(\frac{b}{2},\frac{b}{2})$

Also in second quadrant x is negative and y is positive thus, point S has coordinate $(\frac{-b}{2},\frac{b}{2})$

in third quadrant both x and y is negative thus, point T has coordinate $(\frac{-b}{2},\frac{-b}{2})$

in fourth quadrant y is negative and x is positive thus, point Z has coordinate $(\frac{b}{2},\frac{b}{2})$

Thus, coordinate of given square centered at the Origin, with side length b is W $(\frac{b}{2},\frac{b}{2})$ , S $(\frac{-b}{2},\frac{b}{2})$ , T $(\frac{-b}{2},\frac{-b}{2})$ and Z $(\frac{b}{2},\frac{b}{2})$ .

Thus, option (3) is correct.