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3 years ago

9

Find sin(2x), cos(2x), and tan(2x) from the given information. tan(x) = − 1/2 , cos(x) > 0

1 answer:

vodomira [7]3 years ago

5 0

Answer:

sin(2x) = -4/5
cos(2x) = 3/5
tan(2x) = -4/3

Step-by-step explanation:

It may be easiest to start with tan(2x).

tan(2x) = 2tan(x)/(1 -tan(x)²)

tan(2x) = 2(-1/2)/(1 -(-1/2)²) = -1/(3/4)

tan(2x) = -4/3 . . . . . still a 4th-quadrant angle

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Then cosine can be found from ...

cos(2x) = 1/√(tan(2x)² +1) = 1/√((-4/3)²+1) = √(9/25)

cos(2x) = 3/5

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Sine can be found from these two:

sin(2x) = cos(2x)tan(2x) = (3/5)(-4/3)

sin(2x) = -4/5

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(3x² + 6x + 1) = (3x)

torisob [31]

Answer:

Step-by-step explanation:

x=- $\frac{3-i\sqrt{3} }{6}$ , - $\frac{3+i\sqrt{3} }{6}$

hope this helps! :)

6 0

2 years ago

In a class of 18 students, 5 have a brother and 11 have a sister. There are 5 students who do not have any siblings. What is the

Sphinxa [80]

I think the images should make sense. If not, just comment, I’ll be here. (It’s only letting me insert one image so I’ll have to exp this part) once you’ve filled the two way table in, you circle the two squares that Apple to the situation. In this case, which ones don’t have a brother? The top right and bottom right fit that bill. Then you divide the number of students that actually have a sister but the two squares (added together) that do t have a brother. So it’d be 5/10 which is .5 for an answer

5 0

2 years ago

Read 2 more answers

10 POINTS<br> Find the surface area of the regular pyramid. Round to the nearest tenth if necessary.

NNADVOKAT [17]

Answer:

$SA=234.8\ yd^2$

Step-by-step explanation:

we know that

The surface area of the regular pyramid is equal to the area of the triangular base plus the area of its three triangular lateral faces

step 1

Find the area of the triangular base

we know that

The triangular base is an equilateral triangle

so

The area applying the law of sines is equal to

$A=\frac{1}{2}(14^2)sin(60^o)$

$A=\frac{1}{2}(196)\frac{\sqrt{3}}{2}$

$A=49\sqrt{3}=84.87\ yd^2$

step 2

Find the area of its three triangular lateral faces

$A=3[\frac{1}{2}bh]$

we have

$b=14\ yd$

Find the height of triangles

Applying the Pythagorean Theorem

$10^2=(14/2)^2+h^2$

solve for h

$100=49+h^2$

$h^2=100-49$

$h=\sqrt{51}\ yd$

substitute

$A=3[\frac{1}{2}(14)\sqrt{51}]$

$A=149.97\ yd^2$

step 3

Find the surface area

Adds the areas

$SA=84.87+149.97=234.84\ yd^2$

Round to the nearest tenth

$SA=234.8\ yd^2$

5 0

3 years ago

Read 2 more answers

I forgot how to convert the inequalities so if someone could please help me that would be great!

jeyben [28]

WHEN GRAPHING

Step 1. Convert to y = mx + b

Equation 1: y = 3x/2 - 16/2
Equation 2: y = -5 - 2x

Step 2. Graph like normal

Find zeros & plot
Find y-intercept & plot

Step 3. Shade as indicated by the inequality symbol

> or ≥ = above line
< or ≤ = below line

Step 4. If ≤ or ≥ ONLY, then also shade the line

FOR THIS PROBLEM

1. Graph each equation

2. Shade ABOVE line for each

3. Shade line first equation as well

Hope this helps and God bless!

4 0

2 years ago

Two boys earn money mowing lawns. Jacob mowed 12 lawns this week. He mowed

iris [78.8K]

B is correct
12/3=4
-> 3 x 4 = 12

6 0

2 years ago

Read 2 more answers