The dimensions i.e. length and width of the deck in the drawing is 8 and 6.4 inches respectively.
Given that the length of the previous deck = 15 feet
The width of the previous deck = 12 feet
Since the new deck will add 5 feet to the length and 4 feet to the width,
The length of the new deck = 15 + 5 = 20 feet
The width of the new deck = 12 + 4 = 16 feet
Also given that a drawing of the new deck uses a scale of 1 inch = 2.5 feet.
So, The length of the deck in the drawing = 20/2.5 inches = 8 inches
The width of the deck in the drawing = 16/2.5 inches = 6.4 inches
Therefore, the dimensions i.e. length and width of the deck in the drawing is 8 and 6.4 inches respectively.
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1)6 2)5 3)5 because math is easy hope it helped
Answer:
5h + 3p
Step-by-step explanation:
1 hardback weighs 5 pounds, then
h hardbacks weigh 5 × h = 5h
1 paperback weighs 3 pounds, then
p paperbacks weigh 3 × p = 3p
total weight = 5h + 3p
Let the first number be represented by "x".
Let the second number be represented by "y".
"Four times a number..." = 4x
"... minus 5..." = 4x - 5
"... times another number..." = y(4x - 5)
"... is equal to 21." = y(4x - 5) = 21
y(4x - 5) = 21
"Three times the sum of the two numbers (x & y) is 36." = 3(x +y) = 36
3(x + y) = 36
y(4x - 5) = 21
1) solve for "y"
1) y(4x - 5) = 21
2) 4xy - 5y = 21
3) 4xy/4x - 5y = 21/4x
(what you do to one side of the equals sign, you must do to the other so 4x cancels out on the left side, and on the right side you have to divide 21 by 4x)
4) y - 5y/-5 = 5.25x/-5
5) y + y = -1.05x
6) 2y = -1.05x
7) 2y/2 = -1.05x/2
8) y = -.525x
2) substitute "y" in for the second equation
1) 3(x + (-.525x)) = 36
(notice how you now only have x's in this equation)
3) solve for x
1) 3x - 1.575x = 36
2) 1.425x = 36
3) 1.425x/1.425 = 36/1.425
4) x = 25.3
4) finish finding "y" in either one of the equations using the answer you got for "x"
1) y(4(25.3) - 5) = 21
2) y(101.2 - 5) = 21
3) y(96.2) = 21
4) 96.2y = 21
5) 96.2y/96.2 = 21/96.2
6) y = .22
5) check your work by substituting the two answers back into one of the equations
1) .22(4(25.3) - 5) = 21
2) .22(101.2 - 5) = 21
3) .22(96.2) = 21
4) 21.164 = 21
(round down to get the answer)
Answer:
Max = 86; min = 36.54
Step-by-step explanation:
Step 1. Find the critical points.
(a) Take the derivative of the function.
Set it to zero and solve.
![\begin{array}{rcl}2x - \dfrac{85}{x^{2}} & = & 0\\\\2x^{3} - 85 & = & 0\\2x^{3} & = & 85\\\\x^{3} & = &\dfrac{85}{2}\\\\x & = & \sqrt [3]{\dfrac{85}{2}}\\\\& \approx & 3.490\\\end{array}\](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D2x%20-%20%5Cdfrac%7B85%7D%7Bx%5E%7B2%7D%7D%20%26%20%3D%20%26%200%5C%5C%5C%5C2x%5E%7B3%7D%20-%2085%20%26%20%3D%20%26%200%5C%5C2x%5E%7B3%7D%20%26%20%3D%20%26%2085%5C%5C%5C%5Cx%5E%7B3%7D%20%26%20%3D%20%26%5Cdfrac%7B85%7D%7B2%7D%5C%5C%5C%5Cx%20%26%20%3D%20%26%20%5Csqrt%20%5B3%5D%7B%5Cdfrac%7B85%7D%7B2%7D%7D%5C%5C%5C%5C%26%20%5Capprox%20%26%203.490%5C%5C%5Cend%7Barray%7D%5C)
(b) Calculate ƒ(x) at the critical point.
Step 2. Calculate ƒ(x) at the endpoints of the interval
Step 3.Identify the maxima and minima.
ƒ(x) achieves its absolute maximum of 86 at x = 1 and its absolute minimum of 36.54 at x = 3.490
The figure below shows the graph of ƒ(x) from x = 1 to x = 5.
