Take a look at the attachment below. It fills in for the attachment that wasn't provided in the question -
An oblique pyramid is one that has a top not aligned with the base. Due to this, the height of the pyramid connects with two vertices at its ends to form a right angle present outside the pyramid, knowing that it is always perpendicular to the base. There is no difference between the calculations of the volume of an oblique pyramid and a pyramid however -
![\\Base Area = 2 cm * 2 cm = 4 cm^2,\\Volume ( Pyramid ) = 1 / 3 * ( Base Area ) * ( Height ),\\Volume = 1 / 3 * ( 4 ) * ( 3.75 ),\\-------------------------\\Volume = 5 cm^3](https://tex.z-dn.net/?f=%5C%5CBase%20Area%20%3D%202%20cm%20%2A%202%20cm%20%3D%204%20cm%5E2%2C%5C%5CVolume%20%28%20Pyramid%20%29%20%3D%201%20%2F%203%20%2A%20%28%20Base%20Area%20%29%20%2A%20%28%20Height%20%29%2C%5C%5CVolume%20%3D%201%20%2F%203%20%2A%20%28%204%20%29%20%2A%20%28%203.75%20%29%2C%5C%5C-------------------------%5C%5CVolume%20%3D%205%20cm%5E3)
<u><em>And thus, you're solution is 5 cm^3, or in other words option b!</em></u>
Answer:
It represents an infinite cylinder of radius 4.
Step-by-step explanation:
The first thing to notice is that
![y^2+z^2=4^2](https://tex.z-dn.net/?f=y%5E2%2Bz%5E2%3D4%5E2)
<u>represents a circle of radius 4</u>, with its center in the origin of a plane yz, of cartesians coordinates.
Starting from here, we have to put the coordinate x, for all values of x, to complete the space R³. <em>This will enlarge this circle we had on the plane, to infinity</em> (positive and negative on the x-axis).
Finally, we have that this region is a cylinder of radius 4, with center in y=0 and z=0, and of infinite length in the x coordinates.
How much after subtraction
This is a linear function. Let diameter (inches) be x, circumference (inches) be y, observe that the value of y/x is always 3.14. For example, 15.7/5=31.4/10=...=3.14. Therefore, this is not only a function (one to one correspondence from x to y), it's also a linear function that can be represented as y=3.14x.
Hello,
The formula for finding the area of a circular region is:
![A= \frac{ \alpha *r^{2} }{2}](https://tex.z-dn.net/?f=A%3D%20%20%5Cfrac%7B%20%5Calpha%20%2Ar%5E%7B2%7D%20%7D%7B2%7D%20)
then:
![A_{1} = \frac{80*r^{2} }{2}](https://tex.z-dn.net/?f=A_%7B1%7D%20%3D%20%5Cfrac%7B80%2Ar%5E%7B2%7D%20%7D%7B2%7D%20)
With the two radius it is formed an isosceles triangle, so, we must obtain its area, but first we obtain the height and the base.
![cos(40)= \frac{h}{r} \\ \\ h= r*cos(40)\\ \\ \\ sen(40)= \frac{b}{r} \\ \\ b=r*sen(40)](https://tex.z-dn.net/?f=cos%2840%29%3D%20%5Cfrac%7Bh%7D%7Br%7D%20%20%5C%5C%20%20%5C%5C%20h%3D%20r%2Acos%2840%29%5C%5C%20%5C%5C%20%5C%5C%20sen%2840%29%3D%20%5Cfrac%7Bb%7D%7Br%7D%20%5C%5C%20%5C%5C%20b%3Dr%2Asen%2840%29)
Now we can find its area:
![A_{2}=2* \frac{b*h}{2} \\ \\ A_{2}= [r*sen(40)][r*cos(40)]\\ \\A_{2}= r^{2}*sen(40)*cos(40)](https://tex.z-dn.net/?f=A_%7B2%7D%3D2%2A%20%5Cfrac%7Bb%2Ah%7D%7B2%7D%20%20%5C%5C%20%20%5C%5C%20A_%7B2%7D%3D%20%5Br%2Asen%2840%29%5D%5Br%2Acos%2840%29%5D%5C%5C%20%20%5C%5CA_%7B2%7D%3D%20r%5E%7B2%7D%2Asen%2840%29%2Acos%2840%29)
The subtraction of the two areas is 100cm^2, then:
Answer: r= 1.59cm