Let U= {1,2,3,4,5,6,7,8,9,10}, A={2,4,6,8} B={1,2,3,4}, C={5,6,9}
user100 [1]
Answer: 13. is c, 14. is yes and 15 is no
Step-by-step explanation: Try your best and you might succeed
If two shapes are 'similar', it means that the dimensions of the shapes have the same ratio towards each other. In this example, we can see that the side with a length of 2 correlates to the side with a length of 4. That means, the scaling ratio is 4:2, which can be reduced to 2:1.
The next step is to look at the other side. We know that the width of the smaller shape is 1, and that the second shape is twice the size of the smaller shape (which is what a scale of 2:1 means). Therefore, you just need to multiply the known width by two to get the unknown.
1 * 2 = 2
The missing side _d_ has a length of 2.
Hope that helped =)
Answer: Vertex
Step-by-step explanation:
To solve this exercise you must keep on mind the definitions shown below:
1. By definition, a regular pyramid is a right pyramid whose base is a regular polygon.
2. By definition, a regular polygon is a polygon whose sides have equal lenghts.
3. By definition the apex (which is the vertex at the tip of the pyramid) of a right pyramid lies directly above the center of the base.
Therefore, keeping the above on mind, you can conclude that:
A regular pyramid has a regular polygon base and a Vertex over the center of the base.
Answer:
V = (About) 22.2, Graph = First graph/Graph in the attachment
Step-by-step explanation:
Remember that in all these cases, we have a specified method to use, the washer method, disk method, and the cylindrical shell method. Keep in mind that the washer and disk method are one in the same, but I feel that the disk method is better as it avoids splitting the integral into two, and rewriting the curves. Here we will go with the disk method.
![\mathrm{V\:=\:\pi \int _a^b\left(r\right)^2dy\:},\\\mathrm{V\:=\:\int _1^3\:\pi \left[\left(1+\frac{2}{y}\right)^2-1\right]dy}](https://tex.z-dn.net/?f=%5Cmathrm%7BV%5C%3A%3D%5C%3A%5Cpi%20%5Cint%20_a%5Eb%5Cleft%28r%5Cright%29%5E2dy%5C%3A%7D%2C%5C%5C%5Cmathrm%7BV%5C%3A%3D%5C%3A%5Cint%20_1%5E3%5C%3A%5Cpi%20%5Cleft%5B%5Cleft%281%2B%5Cfrac%7B2%7D%7By%7D%5Cright%29%5E2-1%5Cright%5Ddy%7D)
The plus 1 in '1 + 2/x' is shifting this graph up from where it is rotating, but the negative 1 is subtracting the area between the y-axis and the shaded region, so that when it's flipped around, it becomes a washer.
![V\:=\:\int _1^3\:\pi \left[\left(1+\frac{2}{y}\right)^2-1\right]dy,\\\\\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx\\=\pi \cdot \int _1^3\left(1+\frac{2}{y}\right)^2-1dy\\\\\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\= \pi \left(\int _1^3\left(1+\frac{2}{y}\right)^2dy-\int _1^31dy\right)\\\\](https://tex.z-dn.net/?f=V%5C%3A%3D%5C%3A%5Cint%20_1%5E3%5C%3A%5Cpi%20%5Cleft%5B%5Cleft%281%2B%5Cfrac%7B2%7D%7By%7D%5Cright%29%5E2-1%5Cright%5Ddy%2C%5C%5C%5C%5C%5Cmathrm%7BTake%5C%3Athe%5C%3Aconstant%5C%3Aout%7D%3A%5Cquad%20%5Cint%20a%5Ccdot%20f%5Cleft%28x%5Cright%29dx%3Da%5Ccdot%20%5Cint%20f%5Cleft%28x%5Cright%29dx%5C%5C%3D%5Cpi%20%5Ccdot%20%5Cint%20_1%5E3%5Cleft%281%2B%5Cfrac%7B2%7D%7By%7D%5Cright%29%5E2-1dy%5C%5C%5C%5C%5Cmathrm%7BApply%5C%3Athe%5C%3ASum%5C%3ARule%7D%3A%5Cquad%20%5Cint%20f%5Cleft%28x%5Cright%29%5Cpm%20g%5Cleft%28x%5Cright%29dx%3D%5Cint%20f%5Cleft%28x%5Cright%29dx%5Cpm%20%5Cint%20g%5Cleft%28x%5Cright%29dx%5C%5C%3D%20%5Cpi%20%5Cleft%28%5Cint%20_1%5E3%5Cleft%281%2B%5Cfrac%7B2%7D%7By%7D%5Cright%29%5E2dy-%5Cint%20_1%5E31dy%5Cright%29%5C%5C%5C%5C)
Our exact solution will be V = π(4In(3) + 8/3). In decimal form it will be about 22.2 however. Try both solution if you like, but it would be better to use 22.2. Your graph will just be a plot under the curve y = 2/x, the first graph.
(sqrt7)^6x = 49x - 6
There is no easy way to solve this . Drawing a graph will give a good approximation.
Draw y = (sqrt7)66x
and y = 49x - 6 and see where they intersect
Ill draw it on desmos graphing tool and give you the link.
