Let
x---------> length original square side
x1--------> length resulting rectangle
y1--------> wide resulting rectangle
A1--------> area resulting rectangle
we know that
x1=x+2
y1=x-2
A1=60 in²
60=(x+2)*(x-2)--------> 60=x²-2x+2x-4-------> 60=x²-4------> x²=64
x1=8
x2=-8
the solution is
x=8 in
<span>the area of the original square is 8*8-------> 64 in</span>²
the answer is 64 in²
Answer:
Option 3, p(0) = -10
Option 4, p(10) = 0
Step-by-step explanation:
<u>Step 1: Check</u>
x - 10 + 10 = 0 + 10
<em>x = 10</em>
f(0) = 0 - 10 = -10
x = -10
<em>p(10) = 0</em>
Answer: Option 3, p(0) = -10, Option 4, p(10) = 0
p = 2(l+w)
divide by 2
p/2 = l+w
subtract l from both sides
p/2 - l = 2
Choice c
Answer:
<h3>
41</h3>
Step-by-step explanation:
The consecutive integers are increasing by 1
x ← the smallest integer
x+1 ← the middle integer
x+1+1 = x+2 ← the largest integer
x + x+1 + x+2 = 120
3x + 3 = 120
÷3 ÷3
x + 1 = 40
-1 -1
x = 39
x+2 = 39 + 2 = 41
<span>3x(x^2+4)
= 3x^3 + 12x
hope it helps</span>
