CalculatorUser

I knew my mistake lol

I just wanted some tips on how NOT to make mistakes

Sorry I worded my title wrong

After contemplating about this, I made a list of things to do to avoid silly errors (this may help others too)

- No "bullrushing", don't rush your answer once you have an idea, think it through.

- "Mental Check" Different from double checking, this a method where you mentally check if every step in your solution is practical, makes sense, and mathematicall is right (if you can) AS YOU DO IT

- Ask your self the question: "What is the 'problem' of the Problem?". What is the answer you are trying to find?

- Check for traps. For example, a problem might have "label the units" at the end of it, and you might not notice. These are "traps" that can be avoided.

-Neat and tidy. When doing work on paper, keep everything clear so its easier to do the "mental check" and avoids silly reading and calculation errors.

The solution in the link is not exactly like question you asked, so that's why I pulled out a quote that answers your question correctly.

Here is how u solve this (i solved this 2 days ago so the memory is fresh):

Set up equation - 

\(x+y+z=12\)        (total socks)

\(x+3y+4z=24\)   (total money)

X is $1 socks, Y is $3 socks, and Z is $4 socks.

Use elimination to rid of X. - 

\(2y+3z=12\)

Now guess and check the solutions for (y,z).

After guessing and checking, we find out that the only solution is (3,2).

So we substitute it back into our original equation \(x+y+z=12\)

And solve

\(x+3+2=12\)

\(x=7\)

Answer should be \(\boxed{\text{(D) }7}\)

Post it in latex, I am only going to answer if it is in latex.

If you dunno how to do latex, then remove those extra latex wrappings

Draw two lines

there is one point of intersection

Draw 3 lines, there is 3 points of intersections

Draw 4 lines, there is 6 points of intersections

Draw 5 lines, there is 10 points of intersection

List:

Don't trust me, I might've gotten it wrong.

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:

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

\(\boxed{72000}\)