Part I
We have the size of the sheet of cardboard and we'll use the variable "x" to represent the length of the cuts. For any given cut, the available distance is reduced by twice the length of the cut. So we can create the following equations for length, width, and height.
width: w = 12 - 2x
length: l = 18 - 2x
height: h = x
Part II
v = l * w * h
v = (18 - 2x)(12 - 2x)x
v = (216 - 36x - 24x + 4x^2)x
v = (216 - 60x + 4x^2)x
v = 216x - 60x^2 + 4x^3
v = 4x^3 - 60x^2 + 216x
Part III
The length of the cut has to be greater than 0 and less than half the length of the smallest dimension of the cardboard (after all, there has to be something left over after cutting out the corners). So 0 < x < 6
Let's try to figure out an x that gives a volume of 224 in^3. Since this is high school math, it's unlikely that you've been taught how to handle cubic equations, so let's instead look at integer values of x. If we use a value of 1, we get a volume of:
v = 4x^3 - 60x^2 + 216x
v = 4*1^3 - 60*1^2 + 216*1
v = 4*1 - 60*1 + 216
v = 4 - 60 + 216
v = 160
Too small, so let's try 2.
v = 4x^3 - 60x^2 + 216x
v = 4*2^3 - 60*2^2 + 216*2
v = 4*8 - 60*4 + 216*2
v = 32 - 240 + 432
v = 224
And that's the desired volume.
So let's choose a value of x=2.
Reason?
It meets the inequality of 0 < x < 6 and it also gives the desired volume of 224 cubic inches.
Answer:
14) 21.136505024965 m
15) 43.076759215619 km
Step-by-step explanation:
14)
let h be the height of the lighthouse:
tan(41.3) = 18.7/h
Then
h = 18.7÷tan(41.5)
= 21.136505024965 m
………………………………………
15)
90-71.4 =18.6
Let s be the distance that the plane flew
towards the south:
tan(90 - 71.4) = s/128
Then
tan(18.6) = s/128
Then
s = tan(18.6)×128
= 43.076759215619 km
Answer:
It’s b
Step-by-step explanation:
Answer:
The area of the doghouse is ![32.5\ ft^{2}](https://tex.z-dn.net/?f=32.5%5C%20ft%5E%7B2%7D)
Step-by-step explanation:
we know that
The area of trapezoid is equal to
![A=\frac{1}{2}(b1+b2)h](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B1%7D%7B2%7D%28b1%2Bb2%29h)
where
b1,b2 are the parallel bases of trapezoid
h is the height of trapezoid
in this problem we have
![b1=5\ ft](https://tex.z-dn.net/?f=b1%3D5%5C%20ft)
![b2=5+3=8\ ft](https://tex.z-dn.net/?f=b2%3D5%2B3%3D8%5C%20ft)
![h=5\ ft](https://tex.z-dn.net/?f=h%3D5%5C%20ft)
substitute
![A=\frac{1}{2}(5+8)5=32.5\ ft^{2}](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B1%7D%7B2%7D%285%2B8%295%3D32.5%5C%20ft%5E%7B2%7D)
Answer:
wow thats a lot let me write this down