tertre

We can set up an equation: $(33-x)+(24-x)+x+8=50$ . Solving, we get $57-x+8=50, 57-x=42$ , so   $\boxed{x=15}.$

Assuming that you are meaning cents, we first change $\$4$ into cents, which is $400$ cents.

Now, for the division, $\frac{400}{5}=\boxed{80}$ times/

A quicker way is to use Vieta's, where the sum of the roots are $\frac{-b}{a}$ and the product of the roots are $\frac{c}{a}.$

In the expression $3x^2-4x-7,$   we attain $a=3, b=-4$ , and $c=-7.$

Now, by plugging in our values, the sum of the roots is $\frac{-(-4)}{3}=\frac{4}{3}.$

Consequently, the product of the roots is $\frac{-7}{3}.$

Hope this helped!

Here's a graph on EP's response: 

I might have to change it, took quite a while!

Ugh, my Latex is not working well. 

We see that there is a price of $(70-(t-60))$ for each of the tickets, because there a price of 70 minus 1 for each ticket after the 60th ticket. So the total price is  $(130-t)(t)$and we want this to be greater than $4200$. Rewriting this, we see that $t^2-130t+4200<0.$ This means that $(t-65)^2<25$, so \(-5 Thus \(60  , and the largest possible  $t$ is  $\boxed{69}.$.

This is also distinguishability with duplicates, so $4^4=\boxed{256}.$

See here: https://web2.0calc.com/questions/hlelp-plzzz

We can start off by using the finding angle B. We have $\frac{180-42}{2}=\frac{138}{2}=69^\circ$ . Now, by the tangent formula, the arc AC is twice angle B, so $69*2=138^\circ$ . Now, finally the chord and tangent, we have $\frac{1}{2}*138^\circ=\boxed{69^\circ}.$

Isn't this an angle formed by a tangent and chord. If so, we have angle B and C as 180-42=138 divided by 2 equals 69. Since arc AC has the twice angle measure of angle B, we get 138. Learn this, pretty useful. Thus, angle ACD is 69 degrees.