The complete question is
<span>
Bradley made a house for his dog, Bowser, out of wood with a cube base and a triangular prism top. The dimensions of the dog house are a = 2 feet, b = 1 foot, and c = 1.4 feet.</span><span>
If Bradley plans to paint the outside of the dog house blue, not including the bottom, how many square feet of paint will he use?the picture in the attached figure
</span>
we know that
surface area of the figure=surface area of cube+surface area of triangular prism
surface area of cube=4*a²----> 4*2²-----> 16 ft²
surface area of triangular prism=2*[a*c]+2*[a*b/2]---> 2*[2*1.4]+[2*1]
surface area of triangular prism=5.6+2----> 7.6 ft²
surface area of the figure=16 ft²+7.6 ft²----> 23.6 ft²
the answer is23.6 ft²
Answer:
answer 1 = true
answer 2 . Area of shaded part is 87m^2
<em><u>explanation in the pic above</u></em>
Answer:
$105.48
Step-by-step explanation:
Multiply 131.85 by 0.2 to get 20% of 131.85.
You will get 26.37.
Since it is 4 people, multiply by 4.
You will get $105.48.
The answer is True hope this helps
<span>Length = 1200, width = 600
First, let's create an equation for the area based upon the length. Since we have a total of 2400 feet of fence and only need to fence three sides of the region, we can define the width based upon the length as:
W = (2400 - L)/2
And area is:
A = LW
Substitute the equation for width, giving:
A = LW
A = L(2400 - L)/2
And expand:
A = (2400L - L^2)/2
A = 1200L - (1/2)L^2
Now the easiest way of solving for the maximum area is to calculate the first derivative of the expression above, and solve for where it's value is 0. But since this is supposedly a high school problem, and the expression we have is a simple quadratic equation, we can solve it without using any calculus. Let's first use the quadratic formula with A=-1/2, B=1200, and C=0 and get the 2 roots which are 0 and 2400. Then we'll pick a point midway between those two which is (0 + 2400)/2 = 1200. And that should be your answer. But let's verify that by using the value (1200+e) and expand the equation to see what happens:
A = 1200L - (1/2)L^2
A = 1200(1200+e) - (1/2)(1200+e)^2
A = 1440000+1200e - (1/2)(1440000 + 2400e + e^2)
A = 1440000+1200e - (720000 + 1200e + (1/2)e^2)
A = 1440000+1200e - 720000 - 1200e - (1/2)e^2
A = 720000 - (1/2)e^2
And notice that the only e terms is -(1/2)e^2. ANY non-zero value of e will cause this term to be non-zero and negative meaning that the total area will be reduced. Therefore the value of 1200 for the length is the best possible length that will get the maximum possible area.</span>