#1 If the sides are x and y, then
x^2+y^2 = 68^2 (not √68)
xy = 16
so, y = 16/x
x^2 + 256/x^2 = 68^2
x^4 - 68^2 x^2 + 256 = 0
since 256 = 4*64, we have
(x^2-4)(x^2-64)
x = 2 or 8 (discard the negative roots)
#2
If the cardboard is of side x, then
(x-8)(x-8)4 = 784
x^2 - 16x + 64 = 196
(x+6)(x-22) = 0
x = 22
The box is 14x14x4 = 784 in^3
#3
2x^2+17x+2=0
2x^2 + 17x = -2
x^2 + 17/2 x = -1
x^2 + 17/2 x + (17/4)^2 = -1 + (17/4)^2
(x + 17/4) = 273/16
x = -17/4 ± √273/4
x = 1/4 (-17±√273)
I have about 3 homework problem that I am stuck on! please help me.
#1:
The area of a rectangle is 16, and its diagonal is \sqrt{68}. Find its dimensions and perimeter.
(x^2)+(y^2)=\sqrt{68}
(x^2)+(y^2)=8.246
xy=16
y=16/x
(x^2)+(16/x)^2=8.246
(x^4)+(96/x)=8.246
Then I got stuck please help me continue.
#2:A box with a square base and no top is to be made from a square piece of carboard by cutting 4 in. squares from each corner and folding up the sides. The box is to hold 784 in{}^3. How big a piece of cardboard is needed?
not sure how to approach this
Lastly, #3
2x^2+17x+2=0
x^2+17/2 x+ 1=0
x^2+17/2 x= -1
I know to get my final answer I must complete the square but I am not sure how to do so. This is a far as i got before getting stuck.
1 answer
1 answer