Answer:
150 student tickets and 375 adult tickets
Step-by-step explanation:
Let x = # of student tickets sold, and y = # of adult tickets sold.
Total of 525 tickets were sold, so the first equation is x + y = 525. The register gained $8325, so 10.50x + 18y = 8235 is the second equation.
x + y = 525
10.50x + 18y = 8325
18x + 18y = 9450
10.50x + 18y = 8325
7.5x = 1125
x = 150
150 + y = 525
y = 375
Answer:
Quadratic Equation | Factoring
Solve each quadratic equation using factoring.
1) v² + 5v + 6 = 0
doing middle term factorisation
v²+(3+2)v+6=0
v²+3v+2v+6=0
v(v+3)+2(v+3)=0
(v+3)(v+2)=0
either
<u>v=-3</u>
<u>or</u>
<u>v=-2</u>
2) g² - 3g = 4
keeping all terms in one side
g²-3g-4=0
doing middle term factorisation
g²-(4-1)g-4=0
g²-4g+g-4=0
g(g-4)+1(g-4)=0
(g-4)(g+1)=0
either
<u>g=4</u>
<u>or</u>
<u>g=-1</u>
3)w² + 4w = 0
w(w+4)=0
either
<u>w=0</u>
<u>or</u>
<u>w=-4</u>
4) s² - 8s + 12 = 0
doing middle term factorisation
s²-(6+2)+12=0
s²-6s-2s+12=0
s(s-6)-2(s-6)=0
(s-6)(s-2)=0
either
<u>s=6</u>
<u>or</u>
<u>s=2</u>
5) x ²+ 2x - 35 = 0
doing middle term factorisation
x²+(7-5)x-35=0
x²+7x-5x-35=0
x(x+7)-5(x+7)=0
(x+7)(x-5)=0
either
<u>x=-7</u>
<u>or</u>
<u>x=5</u>
6) r(r + 2) = 99
opening bracket
r²+2r=99
keeping all terms in one side
r²+2r-99=0
r²+(11-9)r-99=0
r²+11r-9r-99=0
r(r+11)-9(r+11)=0
(r+11)(r-9)=0
either
<u>r=-11</u>
<u>or</u>
<u>r=9</u>
7)k(k-4)=-3
opening bracket
k²-4k=-3
keeping all terms in one side
k²-4k+3=0
k²-(3+1)k+3=0
k²-3k-k+3=0
k(k-3)-1(k-3)=0
(k-3)(k-1)=0
either
k=3
or
k=1
8)t²+ 3t + 2 = 0
doing middle term factorisation
t²+(2+1)t+2=0
t²+2t+t+2=0
t(t+2)+1(t+2)=0
(t+2)(t+1)=0
either
<u>t</u><u>=</u><u>-</u><u>2</u>
<u>or</u>
<u>t</u><u>=</u><u>-</u><u>1</u>
9)m ^ 2 - 81 = 0
m²=81
doing square root in both side
<u>m=±9</u>
<u>either</u>
<u>m</u><u>=</u><u>9</u>
<u>or</u>
<u>m</u><u>=</u><u>-</u><u>9</u>
10) h²- 17h + 70 = 0
doing middle term factorisation
h²-(10+7)h+70=0
h²-10h-7h+70=0
h(h-10)-7(h-10)=0
(h-10)(h-7)=0
either
<u>h</u><u>=</u><u>1</u><u>0</u>
<u>or</u>
<u>h</u><u>=</u><u>7</u>
It is true because a rational function is defined as those functions where the variable is placed in the denominator, which must be restricted, because all denominators cannot be equal to zero, other wise it would be undetermined