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:
(4c-1)(y+3)
Step-by-step explanation:
4c(y+3) -(y+3)
=(4c-1)(y+3)
The answer would be 1437 due to the zero property. Therefore you only add the other two numbers.
Using formula sin^2 x +cos^2 x = 1 you get 1 - cos^2(9x) = sin^2(9x)
so in this way this fraction will be rewrited
sin9x/(1-cos^2(9x) = sin9x / sin^2(9x) = 1/sin9x
Answer:
Step-by-step explanation:
Number of students in her class = x students
x is 2 more than a multiple of 4
This means x is an even number because multiples of 4 are even and adding 2 does not change its parity.
X is also 1 more than a multiple of 5. And we know x is and even number it has to be a multiple of 5 ending in 5, not 0. So 5 + 1 is 6.
All multiples of 4 ending in six are 16, 36, 56 ..... and so on
Since there are 15 girls and lower no of boys we need to find out which number of boys are there in 16, 36, 56.
16- 15 = 1
36 - 15 = 21
And 21 is more than 15 therefore the answer is 16 students. 15 girls and 1 boy.
