Answer:
Step-by-step explanation:
Let the length of the rectangle be L.
Let the width of the rectangle be W.
Given the following data;
Area of rectangle = 108 ft²
L = 2W + 6
To find the length and width of the wall of the barn;
<span>-9=2n
n = -9/2 = -4.5</span>
Answer:
The dimension of the plot is 30 yd by 20 yd
Step-by-step explanation:
Given;
Area = 600 yd^2
Length = width + 10
l = w + 10 ......1
Area of a rectangular plot is;
Area A = length × width
A = l × w
Substituting equation 1;
A = (w+10) × w
A = w^2 + 10w
600 = w^2 + 10w
w^2 +10w -600 = 0
Solving the quadratic equation;
w = -30 or 20
cannot be negative
w = 20 yd
l = w+10 = 20+10 = 30yd
The dimension of the plot is 30 yd by 20 yd
Answer:
−
24
x
−
12
h
+3
Step-by-step explanation:
8x-(12x^2+5x)
<em>difference</em>
<u>f
(
x
+
h
) and f
(
x
)
, and plug these values into the difference quotient formula</u>
−
24
x
−
12
h
+3
Answer:
The common difference is -5/4
T(n) = T(0) - 5n/4,
where T(0) can be any number. d = -5/4
Assuming T(0) = 0, then first term
T(1) = 0 -5/4 = -5/4
Step-by-step explanation:
T(n) = T(0) + n*d
Let
S1 = T(x) + T(x+1) + T(x+2) + T(x+3) = 4*T(0) + (x + x+1 + x+2 + x+3)d = 240
S2 = T(x+4) + T(x+5) + T(x+6) + T(x+7) = 4*T(0) + (x+5 + x+6 + x+7 + x+8)d = 220
S2 - S1
= 4*T(0) + (x+5 + x+6 + x+7 + x+8)d - (4*T(0) + (x+1 + x+2 + x+3 + x+4)d)
= (5+6+7+8 - 1 -2-3-4)d
= 4(4)d
= 16d
Since S2=220, S1 = 240
220-240 = 16d
d = -20/16 = -5/4
Since T(0) has not been defined, it could be any number.
