<span>x^2 = 81
x = </span>±√81
x = ±9<span>
x^2 = -9 no solutions
x^2 = 20
x = </span>±√20
x = <span>±2</span>√5<span>
</span>
<span>
</span>
Answer:
10% off would be $96.94
So using the Coupon for 10% off would be the best deal.
Please, for clarity, use " ^ " to denote exponentiation:
Correct format: x^4*y*(4) = y*x^2*(13)
This is an educated guess regarding what you meant to share. Please err on the side of using more parentheses ( ) to show which math operations are to be done first.
Your (x+y)2, better written as (x+y)^2, equals x^2 + 2xy + y^2, when expanded.
The question here is whether you can find this x^2 + 2xy + y^2 in your
"X4y(4) = yx2(13)"
Please lend a hand here. If at all possible obtain an image of the original version of this problem and share it. That's the only way to ensure that your helpers won't have to guess what the problem actually looks like.
Fix you some claculations?
Answer:
x=17/3, y=10/3. (17/3, 10/3).
Step-by-step explanation:
4x-5y=6
2x+5y=-28
-----------------
4x-5y=6
-2(2x+5y)=-2(-28)
--------------------------
4x-5y=6
-4x-10y=56
------------------
-15y=-50
15y=50
y=50/15
y=10/3
4x-5(10/3)=6
4x-50/3=6
4x=6+50/3
4x=18/3+50/3
4x=68/3
x=(68/3)/4
x=(68/3)(1/4)=68/12=17/3
