X+Y=10
XY=-300
Solve for one of the variables
X=10-Y
Plug into the other equation
Y(10-Y)=-300
Now, you have a quadratic...
-y²-10y+300=0
I would multiply everything by -1, so there's no coefficient on y²
y²+10y-300=0
And solve...
-10±√(100-4(-300))
----------------------------
2
Y=-5±5√(13)
Plug back onto original equation...
X+(-5±5√(13))=10
X±5√(13)=15
X=15±5√(13)
Note: There are 2 sets of answers, keep the negatives with the negatives, and the positives with the positives when dealing with the ±.
Answer:
Part A: 18x^2+63x+55, B: second degree trinomial, Part c: (3x+5) (6x+11) =18x^2+63x+55
Answer:
1280
Step-by-step explanation:
rewrite (-8) times (-8) as an equivalent addition problem which is 8 times 8 = 64
take 64 then times it by 20 which is 64 times 20 equals 1,280
Hope this helps:)
Answer:
Step-by-step explanation:
<h3>I believe it is F!</h3><h3 />
Answer:
3 1/3 and 2
Step-by-step explanation:
Let the numbers be x and y and let z be the constant of proportion. Then
xy = 5z ...............(1)
x + y = 4z ...........(2)
x - y = z ..............(3) Adding (2) and (3)
2x = 5z
x = 5z/2 ................(4)
Substituting for x in equation 1:
5z/2 * y = 5z
y = (5z * 2) / 5z = 2.
Substituting for y in equations (2) and (3):
x + 2 = 4z
x - 2 = z Subtracting:
4 = 3z
z = 4/3.
Substituting for z in equation (4)
x = 5(4/3) /2
x = 20/6 = 10/3 = 3 1/3.
So the 2 numbers are 3 1/3 and 2.
Checking the results:
xy = 2 * 10/3 = 20/3 = 5 * 4/3 = 5z
x + y = 10/3 + 2 = 16/3 = 4 * 4/3 = 4z.
x - y = 10/3 - 2 = 4/3 = z.
