MathProblemSolver101

Slope: $\frac{36-30}{23-0} = \frac{6}{23}$

$y = \frac{6}{23} x + b$

$30 = b$

$y = \frac{6}{23} x + 30$

$y \in \mathbb{R}$

For the last box, I don't know the options, so how about you show us the options for the last one.

$\frac{9!}{2!3!4!}$

$x + y + 3z = 10$

$-4x + 3y + 5z = 7$

$kx + z = 3$

Rearranging the 3rd equation, we have $z = 3 - kz$

Let's multiply the 1st equation by $3$: $3x + 3y + 9z = 30$

Subtract the 2nd equation from the new 1st: $7x + 4z = 27$

Plug $3-kz$ for $z:$ $7x + 4(3 - kz) = 27$

Simplifying, we have: $(7-4k)x = 15$

Note: If $7 - 4k = (- \infty, 0) \cup (0, \infty)$ then there will always be a solution for $x,$ and following, $y$ and $z$

Thus, $7-4k$ has to equal $0$ for there to be no solutions for $x,$ which will have no solutions also for $y$ and $z$

We have $7 - 4k = 0$

$4k = 7$

$k = \boxed{\frac{7}{4}}$

Let John's age be $J$

Let his sister's age be $S$

$J = 10S$

$J + 8 = 18$

$J = \boxed{10}$

$f(17) = 17 + 13 = 30$

$f(30) = 30 + 13 = 43$

$f(43) = 43 + 13 = \boxed{56}$

$10p + 8 = 18$

$10p = 10$

$\boxed{p = 1}$

Thats a very nice trick, catmg!

$5^{40} - 5^{39} = 5^{39}(5 - 1) = 5^{39}(4)$

Thanks!