Answer:
p-q= -4
p=q-4
x^2 + px + q=0
=> x^2 + (q-4)x + q = 0
d= b^2-4ac
=(q-4)^2 - 4(1)(q)
= q^2 + 16 -8q - 4q
=q^2 -12q+16
given that D=16
q^2 -12q+16 = 16
=> q^2 -12q = 0
=> q^2 = 12q
=> q = 12q/q
=> q= 12
Therefore the value of q is 12 and the value of p is q-4 = 12-4= 8
Hope you understood............
Write the correct equation. Direct variation problems are solved using the equation y = kx. When dealing with word problems, you should consider using variables other than x and y, you should use variables that are relevant to the problem being solved. Also read the problem carefully to determine if there are any other changes in the direct variation equation, such as squares, cubes, or square roots.
Answer:
Step-by-step explanation:
f(x)= (x - 3)(x - 5)
f(x) = x^2 - 5x - 3x + 15
f(x) = x^2 - 8x + 15
Answer is A
5(3p+2) - 2(5p-3) Distribute the 5 and the 2
(5*3p) + (5*10) + (-2*5p) + (-2*-3)
15p + 50 - 10p + 6 Combine like terms
5p + 56
9514 1404 393
Answer:
Step-by-step explanation:
Let a and s represent the prices of adult and student tickets, respectively.
13a +12s = 211 . . . . . . ticket sales the first day
5a +3s = 65 . . . . . . . ticket sales the second day
Subtracting the first equation from 4 times the second gives ...
4(5a +3s) -(13a +12s) = 4(65) -(211)
7a = 49 . . . . . . . simplify
a = 7 . . . . . . . divide by 7
5(7) +3s = 65 . . . . substitute into the second equation
3s = 30 . . . . . . . subtract 35
s = 10 . . . . . . . divide by 3
The price of one adult ticket is $7; the price of one student ticket is $10.
