Ako180

Alfonso's age=x

Kendra's age=y

y+20=x

(x-7)+(y-7)=54

x+y=68

x-y=20

2x=88

x=44

y+20=44

y=24

Kendra is 24 while Alfonso is 44

Thank you!! That makes a lot of sense

That's what I initially thought but it was incorrect

\(a=\pi r^2\)

\(a=\pi2^2\)

\(a=4\pi\)

\(a=\pi4^2\)

\(a=16\pi\)

\(16\pi-4\pi=12\pi\)

The difference is \(12\pi\)

Hope this helps!!!

-Ako

\(\)4!=24

The factors of 24 are 1,2,3,4,6,8,12,24 and there are 8 of them

Therefore there is an \(\frac{8}{24}\) or \(\frac{1}{3}\) chance that the number chosen is a factor of 4!

Hope this helps!

-Ako

We know that 360 degrees is a full circle

Since \(\frac{72}{360}\) degrees of the circle's yellow

We can simplify that to \(\frac{1}{5}\)

One-fifth of the students favorite color is yellow

That means that \(10\) students favorite color is yellow

Hope this helps!!!
-Ako

We have 3 different combinations we can make

1.) White and Brown: \(4\cdot4=16\) ways

2.) Brown and Blue:\(4\cdot2=8\) ways

3.) White and Blue: \(4\cdot2=8\) ways

So if we add all of the ways we get \(16+8+8=32\) different ways

Hope this helps!

-Ako

There is no diagram. Can you attach it?

We can solve this using a system of equations. 

Let \(x=\)tickets bought at the door

Let \(y=\)tickets bought in advance

Our equations are 

\(10y+15x=1390\)

\(y+x=120\)

We can solve elimination (Feel free to use a different strategy):

Multiply the bottom equation by \(10\)

\(10y+10x=1200\)

Subtract the bottom equation

\(5x=190\)

Divide both sides by \(5\)

\(x=38\)

Plug that value into the bottom equation (or the top if you want)

\(y+38=120\)

Subtract \(38\) from both sides

\(y=82\)

\(82\) tickets were bought at the door.

Hope this helped!!!

-Ako

\( \frac{x}{x}-5=\frac{4}{x}-4\)

We know that \(\frac{x}{x}\) is \(1\), therefore meaning that the left side of the equation is, 

\(1-5=-4\)

Therefore our equation is now

 \(-4=\frac{4}{x}-4\)

From there we can simplify by adding 4 to each side 

\(0=\frac{4}{x}\)

There is no value of \(x\) that satisfies this equation. 

Therefore there are no solutions