Answer:
1. 16
2. -4
Step-by-step explanation:
First factor the expression
lim x→1 (x^3 + 5x^2 + 3x-9)/(x-1)
lim x→1 (x-1) (x+3)^2/(x-1)
Canceling the x-1 in the top and bottom
lim x→1 (x+3)^2
Let x=1
lim x→1 (1+3)^2 = 4^2 = 16
2. lim x→0 (x^2 -6x+8) /(x-2)
First factor the expression
lim x→0 (x-4) (x-2) /(x-2)
Canceling the x-2 in the top and bottom
lim x→0 (x-4)
Let x=0
lim x→0 (0-4) = -4
Filling in the table
x -.1 -.01 -.001 .001 .01 .1
f(x) -4.1 -4.01 -4.001 -3.999 -3.99 -3.9
Answer:

Step-by-step explanation:
We have to find the formula for the value of 'l'.
So, we will isolate 'l' on one side of the formula.
A = 2lw + 2lh + 2wh
A = 2l(w + h) + 2wh
A - 2wh = 2l(w + h) + 2wh - 2wh
A - 2wh = 2l(w + h)


Therefore, formula for 'l' will be,

Answer:
The largest total area that can be enclosed will be a square of length 272 yards.
Step-by-step explanation:
First we get the perimeter of the large rectangular enclosure.
Perimeter of a rectangle =2(l + w)
Perimeter of the large rectangular enclosure= 1088 yard
Therefore:
2(L+W)=1088
The region inside the fence is the area
Area: A = LW
We need to solve the perimeter formula for either the length or width.
2L+ 2W= 1088 yd
2W= 1088– 2L
W = 
W = 544–L
Now substitute W = 544–L into the area formula
A = LW
A = L(544 – L)
A = 544L–L²
Since A is a quadratic expression, we re-write the expression with the exponents in descending order.
A = –L²+544L
Next, we look for the value of the x coordinate


L=272 yards
Plugging L=272 yards into the calculation for area:
A = –L²+544L
A(272)=-272²+544(272)
=73984 square yards
Thus the largest area that could be encompassed would be a square where each side has a length of 272 yards and a width of:
W = 544 – L
= 544 – 272
= 272 yards
Answer: Change the number: 36
To:48
Step-by-step explanation:
12x2 is 24 but 24 x 2 isn’t 36
Answer:
42
Step-by-step explanation:
6x7=42