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Thanks!  I also need help with these problems:

Get lost guest  and put your effort into doing your own homework.

Maybe on Mars.  There are some pretty high ones there. 

_.

The area is 15 + 12*sqrt(2).

1. The set of possible x is [3,25].

2.  The smallest possible perimeter is 23.

3. The largest possible third side is 29.

at 7  y/o he hadd 22 coins      He added  (577-22) = 555 coins over the years

7 yrs   +   555 coins/15 coins/yr   =   7 +  37 yrs = 44 y/o

how would i solve for the adjacent so I can do the next one?

it was correct,thank you!

A = aldo

C = Carlos = 3A

K = Karen = A + 8     Summed they equal 93

A   +   3A     +   A+8   = 93      Solve for A = Aldo          then Carlos = 3 * Aldo           and Karen = Aldo + 8

Hello, Hereuka. Thank you so much for your time and effort to respond to it. I will thouroughly revise your response slowly. Since I don’t know much trigonometry. Once again, thank you.

This is correct in structure then, yes?

There are 1000 grams in a kilogram; so in 1 1/2 kilograms of frosting, she has 1500 grams of frosting.

Since she uses 50 grams of frosting on each cupcake, she can frost  1500 / 50  =  answer  cupcakes.

y = a·sin(b·x) + c

c is the value of the midline; c = 70

a is the amplitude (high or low from the midline); a = 10      (found by high - midline)

b is the modification of the period; 360° in a 24 hour day; b = 360/24.

Since AB || CD, angle(APR) = angle(QRP)              [alternate interior angles]

Since PQ = QR, angle(QRP) = angle(QPR)             [base angles of an isosceles triangle]

Since angle(APR) = x, angle(QRP) = x                    [substitution, or transitivity]

angle(BPQ) + angle(QPR) + angle(RPA) = 180°      [they form a line]

Therefore:  (100° + x) + (x) + (x) = 180°

                               100° + 3x  =  180°

                                           3x  =  80°

                                             x  =  (80/3)°

ABC is an equilateral triangle with side length 4. M is the midpoint of BC, and AM is a diagonal of square ALMN. Find the area of the region common to both ABC and ALMN.

OOPS! That was my fault. I'm so sorry. I'll correct that.

In the problem, you have a "3"; in your answers, you have "13".

The graph will be horizontally compressed by the factor.

Check this out by graphing each one.

Sine wave  amplitude = 2    and inverted (flipped around x-axis) due to the ' - ' sign......looks OK to me !

I think you are correct....it does work against the force of FRICTION, but does not move it vertically agains the force of gravity....

You can pick either of the equations....just put in  -3   where 'y' is to find x

x + 2(-3) = -4

Or do i put -3 in the x

So rather than putting 

 x+2y=-4

I put 

 x+-3y=-4

2x-3y=13

 x+2y=-4       Multiply THIS equation by -2

-2x-4y=8      add this to TOP equation

-7y = 21

   y = -3      now substitue this 'y' value back into one of the equations to find the 'x' value of the coordinate

10 coins to start    5 dimes       5 out of 10 to pick a  dime first     5/10

9 coins left             3 quarters   3 out of 9  to pick a Q                   3/9

5/10  *  3/9  = 15/90 = 1/6   

Start with a sine wave

sin x

Amplitude  3

3 sin x               

Midline y=2    (usual midline is 0....add 2)

3 sin x     + 2

Period 4pi   (usual period is 2 pi....divide x by 2)

3 sin x/2    + 2  

See here :

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