Answer:
RL=5x+28 and
RO=8X-11
diagonal of square bisect equally the side
:.5x+28=8x-11
11+28=8x-5x
39=3x
x=39/3=13
<u>RY</u><u>=</u><u>RL</u><u>=</u><u>5</u><u>x</u><u>+</u><u>2</u><u>8</u><u>=</u><u>5</u><u>×</u><u>1</u><u>3</u><u>+</u><u>2</u><u>8</u><u>=</u><u>9</u><u>3</u><u>If the answer is 93, move to </u><u>answer</u><u>.</u>
The dimensions of the rectangular pen should be 15 by 20 feet and the maximum area is 1200 square feet.
Let the area be y .
Area = (base) × (height)
Base = 2x
Height = h
Let the area of the rectangular pens be y .
∴ y = 2xh
Perimeter of all the fencing = 4x+3h
∴ 4x+3h = 120
now we solve for h
3h = 120-4x
h = 40 - 4/3 x
Now we will substitute this value in the above first equation:
y = 2xh
or, y = 2x (40 - 4/3 x)
or, y = 80x - 8/3 x²
Now for the maximum area we have to find the first order differentiation of y
now,
dy /dx = 80 - 16/3 x
At dy/dx = 0 we get the value of x for which y is maximum.
80 - 16/3 x = 0
or, - 16/3 x = -80
or, x = 15 feet
Hence height = 40 - 4/3 x = 40 - 20 = 20feet
Maximum area = 2xh = 2×15×40 = 1200 square feet
The dimensions of the rectangular pen should be 15 by 20 feet and the maximum area is 1200 square feet.
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A><span><span>−b</span>−<span>4
Hope this helps. c:</span></span>
Answer:
see below
Step-by-step explanation:
The graph of it on a number line is an open circle at x=3 with a line extending to the right through larger numbers.
When the inequality does not include the "or equal to" case, the boundary is graphed as a dashed line (on an x-y plane) or open circle (on a number line). The shaded area covers values of the variable that meet the condition of the inequality. Here, those are values of x that are more than 3.
First we'll do two basic steps. Step 1 is to subtract 18 from both sides. After that, divide both sides by 2 to get x^2 all by itself. Let's do those two steps now
2x^2+18 = 10
2x^2+18-18 = 10-18 <<--- step 1
2x^2 = -8
(2x^2)/2 = -8/2 <<--- step 2
x^2 = -4
At this point, it should be fairly clear there are no solutions. How can we tell? By remembering that x^2 is never negative as long as x is real.
Using the rule that negative times negative is a positive value, it is impossible to square a real numbered value and get a negative result.
For example
2^2 = 2*2 = 4
8^2 = 8*8 = 64
(-10)^2 = (-10)*(-10) = 100
(-14)^2 = (-14)*(-14) = 196
No matter what value we pick, the result is positive. The only exception is that 0^2 = 0 is neither positive nor negative.
So x^2 = -4 has no real solutions. Taking the square root of both sides leads to
x^2 = -4
sqrt(x^2) = sqrt(-4)
|x| = sqrt(4)*sqrt(-1)
|x| = 2*i
x = 2i or x = -2i
which are complex non-real values