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Ohhhhhh ok i see thats ok 

soultion1: Every 30 seconds 3-2=1 toys are put in the box, so after 27\cdot30 seconds, there will be 27 toys in the box. Mia's mom will then put 3 toys into the box and we have our total amount of time to be 27\cdot30+30=840 seconds, which equals 14 minutes

soultion2: Though Mia's mom places 3 toys every 30 seconds, Mia takes out 2 toys right after. Therefore, after 30 seconds, the two have collectively placed 1 toy into the box. Therefore by 13.5 minutes, the two would have placed 27 toys into the box. Therefore, at 14 minutes, the two would have placed 30 toys into the box. Though Mia may take 2 toys out right after, the number of toys in the box first reaches 30 by 14 minutes

She has to pick up 30 toys.

Mia's mom puts 3 toys, and mia takes 2 toys out.

So 3 - 2 = ONE TOY being put into the box EVERY 30 seconds.

Since one toy is being put into the box every 30 seconds.

It will take 30 * 30 = 900 seconds.

900/60 = 15 minutes

oh the burden of being a parent, good thing for me I am still a youngling

literally substitute

\(f(x)=x+2\)

\(g(x)=x^2\)

So first find \(f(g(x))\)

That is \(f(x^2)\)

Then find \(g(f(x))\)

That is \(g(x+2)\).

So now we have \(f(x^2)=g(x+2)\)

Ok now solve

I am not educated on these type of functions, so when I personally solved it, I got:

x^2+2=(x+2)^2 = x=-((1/2))

\(\boxed{x=-\frac{1}{2}}\)

I am not sure someone please check my calculations!

Bro post in rendered latex please! Stop being lazy buddy!
 

Suppose \(f(x)=\frac{3}{2-x}\). If \(g(x)=\frac{1}{f^{-1}(x)}+9\), find \(g(3)\).

Find the inverse of function f(x).     You do this by setting up f(x) as y, then switch "x" and "y". Solve for y.

.

\(y=\frac{3}{2-x}\)

\(x=\frac{3}{2-y}\)

\(2x-yx=3\)

\(2-y=\frac{3}{x}\)

\(y=-\frac{3}{x}+2\)

\(y=2-\frac{3}{x}\)

So now we figured out f^-1(x). We then plug that into g(x).

\(g(x)=\frac{1}{2-\frac{3}{x}}\)

We then evaluate g(3).

I will leave that up to you now

He only uses 4 at a time out of 5 tires.

This question seems hard, but it actually is this simple:

\(\frac{4}{5}*90000\)

\(\boxed{72000}\)

oops! I think the website covers parallelogram law of vectors addition.

...but there was no answer choice...

eh, it makes sense.

thanks CU :D

ill take a look at it rn

Circumradius = 2*30/10 = 6.

The answer was actually none of the above, which makes sense if you draw it out.

This can be set up as a simple proportion.

HEIGHT OF ALICIA : SHADOW OF ALICIA = FLAGPOLE HEIGHT : SHADOW OF FLAGPOLE

Remember, we have to the keep the units consistent.

150 cm : 200 cm = x : 2400 cm 

150 * 2400 = 200x 

x = 150 * 2400/200

x = 150 * 12

x = 1800 cm

In meters, it would be 18 meters.

The answer is 8.8x10^6

√ [ (3 - 0)^2 + (3 - 0)^2 ] = √  [9 + 9 ]  = √18  =  3√2 units

 https://web2.0calc.com/questions/please-help-geometry-problem

2 and 3.

(3) is not the only correct answer, there is more than one.

B

If you draw a graph or look it up on Desmos, it curves down around y = 10. It only hits the second and third quadrant. B is the only situation that describes this. 

ABC must be a right triangle, so statement (3) must be true.

fancy

oops i screwed up my brain is wired wrong!

anyways, try what I did, find the SMALLEST number that fits the problem. and prime factorize it.

I just rounded

The quadratic  evaluates to

[√105 - 1 ]

________    =    sin  θ

     12

So

√3sin  ( arcsin ( [√105 - 1 ]/12 ) )  +  cos (arcsin ( [√105 - 1 ] /12) )  =

√35

 ___   ≈   1.972

   3

Hi CalculatorUser,

Here is the pic

If ΔABC is dilated from point A by a scale factor of one half which of the following equations is true about segment D prime E prime?

3x^2 - 4y^2 + 4x -8y + 4  =  0

3x^2  +y^2  + 4x -3y + 4  = 0

Subtract the first equation from the second and we get that

5y^2  + 5 y   =   0   divide through by  5

y^2 + y  = 0       factor

y ( y + 1)   =   0

Setting   each  factor  to 0   and solving for  y  produces  :

y  =  0      and  y   =  -1

When y  =  0

We have that both equations evaluate to

3x^2 + 4x + 4  = 0    (1)

This has no real solutions for  x   when y  = 0 because the discriminant for (1)  is less than  0

So....there  are no real intersection points  when  y  = 0

When  y  =  -1

The first equation evaluates to

3x^2 - 4 + 4x + 8 + 4  =  0

3x^2 + 4x + 8  = 0       (2)

And the second equation evaluates to

3x^2 + 1 + 4x + 3 + 4  = 0

3x^2 + 4x +  8  =  0    (3)

(2)  and (3)  are the same

Again  when y  =  -1  there are no real solutions for x because  the discriminant is less than 0

So....we  have no  real  solutions   in x  for either y value....

WolframAlpha confirms this   :  https://www.wolframalpha.com/input/?i=3x%5E2+-+4y%5E2+%2B+4x+-8y+%2B+4++%3D+0%2C+++3x%5E2++%2By%5E2++%2B+4x+-3y+%2B+4+%3D+0++

oh, I must of overcomplicated things, thanks!