A) If you cut off half the peaks (3.5 ft down from the top), they will just fill the valleys making a rectangle 18.5 ft by 30 ft. The area is the product of these dimensions.
A = (18.5 ft)*(30 ft) = 555 ft²
b) To determine the number of quarts of paint required, divide this number by 90 ft². Since you cannot buy a partial quart, if there is any remainder, use the next higher integer number of quarts.
555/90 ≈ 6.2 so Pat should buy 7 quarts of paint.
Answer:
34
Step-by-step explanation:
Straight lines are 180 degrees, so the angle inside of the triangle next to the 127, is 53 degrees. (180-127= 53)
So to find "x", we need to figure out what the full number is.
59+53 = 112.
180-112= 68.
Now that we have the degree of the full angle, we can make an equation to find x.
x*2 = 68
(Divide 2 from both sides to isolate the variable)
x = 34
I hope this helped!
It wants it to be in slope-intercept form.
y=mx+b
We have to first find the slope and plug it into point-slope form.
y-y1=m(x-x1)
Find the slope of the second line. (I did this one first on accident)
Rise/run= 3/1= 3 The slope is 3. Plug that in along with the point (0,3)
y-3=3(x-0)
y-3=3x
Add 3 to the other side.
y= 3x +3 <- <em>for the second line</em><em>
</em>
Now, the second.
rise/run= 1/2= .5 Use point (6,0)
y-0=.5(x-6)
y= .5x-3
y=.5x-3 <- for the first line
I hope this helps!
~kaiker
Answer:
26 ft square by 13 ft high
Step-by-step explanation:
The tank will have minimum surface area when opposite sides have the same total area as the square bottom. That is, their height is half their width. This makes the tank half a cube. Said cube would have a volume of ...
2·(8788 ft^3) = (26 ft)^3
The square bottom of the tank is 26 ft square, and its height is 13 ft.
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<em>Solution using derivatives</em>
If x is the side length of the square bottom, the height is 8788/x^2 and the area is ...
x^2 + 4x(8788/x^2) = x^2 +35152/x
The derivative of this is zero when area is minimized:
2x -35152/x^2 = 0
x^3 = 17576 = 26^3 . . . . . multiply by x^2/2, add 17576
x = 26
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As the attached graph shows, a graphing calculator can also provide the solution.
