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.