If you have a graphing calculaator enter the matrix then on the outside of it hit the ^ button and put -1
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First you'd substitute x= y-3 in the first equation.
10(y-3) -10y= 1
10y - 30 -10y= 1
The y's cancel each other out, so you're left with -30= 1. That means there's no solution :)
Y=a(x-1)^2 -17 is the equation of the parabola.
at (0,16), 16=a(0-1)^2 -17. Solve for a = 33.
Equation is then y=33(x-1)^2-17 so when y=0,
(x-1)^2 = 17/33
x-1 = + and - square root of 17/33
x = 1.7177 and 0,2823 as the intercepts
I think it’s gonna be 6 adults for every 42 students not really sure so sorry if I got it wrong
Answer:
Formula to find the area of a triangle is 1/2(bxh). The area of the right triangle is 28.
Step-by-step explanation:
To solve this problem you would plug in the base and height into the formula and follow the order of operations (P.E.M.D.A.S). So the equation would then read 1/2(5x11.2). Once you solve inside the parenthesis, the equation then reads, 1/2(56). Then you solve what half of 56 is, which is 28.
I hoped this helped you!
