Answer:
b
Step-by-step explanation:
it has the largest iqr box.
Do you just want the equation ?
Answer:
Step-by-step explanation:
xy = 300
x + y = 50 Solve for y
y = 50 - x substitute into xy = 300
x(50 - x) = 300 Remove the brackets.
50x - x^2 = 300 Bring the left to the right.
0 = x^2 - 50x + 300
This is a quadratic. It will have 2 solutions.
a=1
b = - 50
c = 300
Put these into the quadratic equation.
It turns out that x has two values -- both plus
x1 = 42.03
x2 = 6.97
x1 + y1 = 50
42.03 + y = 50
y = 50 - 42.03
y = 7,97
(42.03 , 7.97)
====================
x2 + y2 = 50
6.97 + y2 = 50
y2 = 50 - 6.97
y2 = 43.03
(6.97 , 43.03)
<u>Answer:</u>
Factor over complex numbers of
is 
<u>Solution: </u>
From question given that
→ (1)
On substituting
in equation (1),

Taking 2 as a common in above expression,

Rewrite the above expression,
![=2\left[y^{2}+2(y)(9)+9^{2}\right]](https://tex.z-dn.net/?f=%3D2%5Cleft%5By%5E%7B2%7D%2B2%28y%29%289%29%2B9%5E%7B2%7D%5Cright%5D)
![\left[\text { Using }(a+b)^{2}=a^{2}+2 a b+b^{2}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Ctext%20%7B%20Using%20%7D%28a%2Bb%29%5E%7B2%7D%3Da%5E%7B2%7D%2B2%20a%20b%2Bb%5E%7B2%7D%5Cright%5D)
![\left[\text { since } y=x^{2}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Ctext%20%7B%20since%20%7D%20y%3Dx%5E%7B2%7D%5Cright%5D)
![\left[\text { Using } a^{2}+b^{2}=(a+i b)(a-i b), \text { where } i=\sqrt{-1}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Ctext%20%7B%20Using%20%7D%20a%5E%7B2%7D%2Bb%5E%7B2%7D%3D%28a%2Bi%20b%29%28a-i%20b%29%2C%20%5Ctext%20%7B%20where%20%7D%20i%3D%5Csqrt%7B-1%7D%5Cright%5D)
![=2[(x+3 i)(x-3 i)]^{2}](https://tex.z-dn.net/?f=%3D2%5B%28x%2B3%20i%29%28x-3%20i%29%5D%5E%7B2%7D)

Hence Factor over complex numbers of
is 
Plug in the values and see which ones do not fi:-
. A 3(-4) + 2(1) = -10
that fits so it must be B, C or D.