3 feet width is added to both sides .
so the dimensions become 34 feet and 36feet
hence the fencing length required or the perimeter= 2(l+b)
= 2( 34+36)
= 2(70)
= 140feet fence
Answer:
10 small boxes and 12 large boxes
Step-by-step explanation:
Let x = number of large boxes
Let y = number of small boxes
We are told that;
volume of each small box = 6 cubic feet
volume of each large box = 22 cubic feet.
Total volume = 324 ft³
Thus;
6x + 22y = 324 - - - (eq 1)
We are told that 22 boxes of paper were shipped.
Thus; x + y = 22 - - - (eq 2)
Making x the subject in eq 2 gives;
x = 22 - y
Put 22 - y for x in eq 1;
6(22 - y) + 22y = 324
132 - 6y + 22y = 324
16y = 324 - 132
16y = 192
y = 192/16
y = 12
So, x = 22 - 12 = 10
Step-by-step explanation:
solution,
midpoint between A(-1,4) and B(-5,-2)=
(X1 + X2 ÷2, Y1 + Y2 ÷2)
=(-1-5÷2 , 4-2)
=(-6÷2 ,2÷2)
=(-3 , 1)
therefore the midpoint between A(-1,4) and B(-5 ,-2)=(-3,1)
I would be a nice person and answer this, but I’m dumb so sorry 4-4x
Answer:
y=81x^4+648x^3+1944x^2+2592x+1289
Step-by-step explanation:
1. Interchange x and y.
- y=\frac{\sqrt[4]{x+7}}{3}-2
- x=\frac{\sqrt[4]{y+7}}{3}-2
2. Then solve x=\frac{\sqrt[4]{y+7}}{3}-2 for y.
3. y=81x^4+648x^3+1944x^2+2592x+1289
I hope this helps.
