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We have two graphs, one for f(x)=Cos2x and another for g(x)=3Sin2x

for both of them being between 0 - 180 deg.

The question is: Calculate the values for x where f(x) = g(x) if x $\in$ (0;180)

Hello juriemagic!

$f(x)=g(x)\\ cos\ 2x=3sin\ 2x$

$u=2x$

$cos\ u=3sin\ u\\ cos^2\ u =3(1-cos^2\ u)\\ cos^2\ u=3-3cos^2\ u\\ 4cos^2\ u=3\\ cos\ u=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}$

$u=30^{\circ}\\ x=\frac{u}{2}$

$x=15^{\circ}$

  !

F = ma

   = 2 kg * 10 m/s²   = 20 N

Has normal period of 2 pi   so sinx

amplitude is 1/2 of the total wave height (4)    so   2 sin x

shifted down one unit     2 sin x   - 1

school bought   100 * (100 +x )   candies

by brute force   x = 6     106 boxes

(10 000 +100x) mod 14 = 2                     10600 mod 14 = 2

Slope, m , of given line = 2/3     perpindicular slope = - 1/m = -3/2

y-5 = -3/2 (x+6)       re-arrange to y=mx + b form to find correct answer 

idk, I did not check them all, if you know how to do one then you should get them all correct.

This means 'solve for x'

y=3-3x

y-3 = -3x

3-y = 3x

x = (3-y)/3

https://www.desmos.com/calculator/roshd2nmok

Which is 2000

Well you could split it up into squares of 16. Which would be the longer, but easier way.

There are two 8x8 squares.

64+64=128

128+16+16+16+16+16=208

Yay!

The expression simplifies to 1/3.

Total number of faculty member is 210, and there are twice as many lecturers as professors.

If we combine the professors and the lecturers, we get 15. 210/15=14

14*10=140

There are 140 lecturers

Yay!

Is there supposed to be a value for y...?

0.001 is 10^-3 and 2000000 is 2*10^6. Multiplied we have

10^-3*2*10^6

2*10^3.

Find the value of f(3)

Let 1 square be 1 unit squared

5 * 3 = 15

Area of the parallelogram is   15 - ( 2 * 4 + 1 ) = 6 units squared

Celsius: -273

Fahrenheit: -459

for #1

And second one all I know is the mass of a hydrogen atom is 1.67 x 10-27 kg and Boltzmann's constant is k = 1.38 × 10-23 Joule/Kelvin. But I don't know how to solve! SOrry!!

Yay, I love rate problems!

Noon is 12:00 PM. 

So train A leaves at 12:00 PM and drives at 90 mph (I know right? CLASS 5 PASSENGER THAT'S LUXURY). And train B leaves from the same place and drives from the same direction at 10 mph greater than 90mph. In 30 minutes 12-12;30, Train A will go 45 miles and Train B will go 0 miles (Not left yet). In another 30 minutes, a total of 1 hour, Train A will have went 90 miles and train B will go 50. 2 hours, 180 miles VS 150 miles. 4 hours, 360 miles VS 350 miles. And 5 hours will add a total of 10 miles difference, which is 450 vs 450!!! Therefore 12:00+5:00=

17:00 in 24 hour clock

and 5:00 PM in 12 hour clock!

I'm sorry, I don't really understand; the problem is already completed.

Pepperoni=P

Mushrooms=M

Sausage= S

Green Peppers=G

Onions= O

C=cheese

Note that all of these pizzas have cheese.

1.P+M+C=14.67

2.P+M+S+G+C=17.85

3.P+M+S+G+O+C=19.44

Subtract equations 2 from 3 and get Onions=$1.59.

Then you can't find out any more, as this is a pizza estimation problem that has too many variables and I don't eat pizza soooooo...

how were the other ones ?

Yes, it means does not exist.