Answer:
The original length was 41 inches and the original width was 16 inches
Step-by-step explanation:
Let
x ----> the original length of the piece of metal
y ----> the original width of the piece of metal
we know that
When squares with sides 5 in long are cut from the four corners and the flaps are folded upward to form an open box
The dimensions of the box are

The volume of the box is equal to


so

simplify
-----> equation A
Remember that
The piece of metal is 25 in longer than it is wide
so
----> equation B
substitute equation B in equation A

solve for y

Solve the quadratic equation by graphing
using a graphing tool
The solution is y=16
see the attached figure
Find the value of x

therefore
The original length was 41 inches and the original width was 16 inches
Let q be the number of 25 cent coins.
Let d be the number of 10 cent coins.
0.25q+0.10d= 3.95...(1)
q-d=6...(2)
(2)-> q-d= 6
q-d+d= 6+d
q= 6+d...(2a)
(2a)-> (1) 0.25q+0.10d=3.95
0.25(6+d)=0.10d= 3.95
1.95+0.25d+0.10d= 3.95
0.35d= 3.95-1.5
0.35d/0.35= 2.45/0.35
d= 7...(3)
(3)->(2) q-d= 6
q-7= 6
q=6+7
q= 13
There are 13 quarters and 7 dimes.
Answer:
w = -3
Step-by-step explanation:
8w + 12 = -12
use inverse operations:
8w + 12 = -12
-12 -12
8w = -24
/8 /8
w = -3
3.341 to the nearest tenth is 3.3
Perimeter = 2L + 2W
Let x = width and let 2x = length
Perimeter = 66
2x + 2(2x) = 66
2x + 4x = 66 --> 6x = 66 --> x = 66/6 --> x = 11
The width of the rectangle is 11 cm.