8 different styles
4x2=8
If you want to answer future questions like this then you multiply the numbers of the items.
All the transformation performed on a pentagon does not affect on its size or its length
It is not stretched or shrunk
So,
The original and transformed images are congruent
This is a common factor problem.
Pencils come in a pack of 12
Erasers come in a pack of 10
First, break the number into their prime factors(the idea is that we will break the number down into its smallest multiples, which are prime numbers):
10 = 2 * 5
12 = 2 * 2 *3
So now we take the unique multiples of each number, and when we multiply them together, we will get the smallest number that both 10 and 12 can be divided into(this is what the problem is asking for)
We have (2*2*3) that comes from 12, and the only unique number that comes from the 10 is (5)
So now, we multiply:
2*2*3*5=60
However, this isn't exactly out answer. Now we have to divide our answer by the number of each this in the pack to know how many packs to buy.
60/12=5 packs of pencils
60/10= 6 packs of erasers
I hope this helps. Let me know if you have any questions!!
<h2>
Hello!</h2>
The answer is:
The third option:
2.7 times as much.
<h2>
Why?</h2>
To calculate how many more juice will the new can hold, we need to calculate the old can volume to the new can volume.
So, calculating we have:
Old can:
Since the cans have a right cylinder shape, we can calculate their volume using the following formula:
![Volume_{RightCylinder}=Volume_{Can}=\pi r^{2} h](https://tex.z-dn.net/?f=Volume_%7BRightCylinder%7D%3DVolume_%7BCan%7D%3D%5Cpi%20r%5E%7B2%7D%20h)
Where,
![r=radius=\frac{diameter}{2}\\h=height](https://tex.z-dn.net/?f=r%3Dradius%3D%5Cfrac%7Bdiameter%7D%7B2%7D%5C%5Ch%3Dheight)
We are given the old can dimensions:
![radius=\frac{5.2cm}{2}=2.6cm\\\\height=9.4cm](https://tex.z-dn.net/?f=radius%3D%5Cfrac%7B5.2cm%7D%7B2%7D%3D2.6cm%5C%5C%5C%5Cheight%3D9.4cm)
So, calculating the volume, we have:
![Volume_{Can}=\pi *2.6cm^{2} *9.4cm=199.7cm^{3}](https://tex.z-dn.net/?f=Volume_%7BCan%7D%3D%5Cpi%20%2A2.6cm%5E%7B2%7D%20%2A9.4cm%3D199.7cm%5E%7B3%7D)
We have that the volume of the old can is:
![Volume_{Can}=199.7cm^{2}](https://tex.z-dn.net/?f=Volume_%7BCan%7D%3D199.7cm%5E%7B2%7D)
New can:
We are given the new can dimensions, the diameter is increased but the height is the same, so:
![radius=\frac{8.5cm}{2}=4.25cm\\\\height=9.4cm](https://tex.z-dn.net/?f=radius%3D%5Cfrac%7B8.5cm%7D%7B2%7D%3D4.25cm%5C%5C%5C%5Cheight%3D9.4cm)
Calculating we have:
![Volume_{Can}=\pi *4.25cm^{2} *9.4cm=533.40cm^{3}](https://tex.z-dn.net/?f=Volume_%7BCan%7D%3D%5Cpi%20%2A4.25cm%5E%7B2%7D%20%2A9.4cm%3D533.40cm%5E%7B3%7D)
Now, dividing the volume of the new can by the old can volume to know how many times more juice will the new can hold, we have:
![\frac{533.4cm^{3} }{199.7cm^{3}}=2.67=2.7](https://tex.z-dn.net/?f=%5Cfrac%7B533.4cm%5E%7B3%7D%20%7D%7B199.7cm%5E%7B3%7D%7D%3D2.67%3D2.7)
Hence, we have that the new can hold 2.7 more juice than the old can, so, the answer is the third option:
2.7 times as much.
Have a nice day!
Answer:
60 degrees
Step-by-step explanation:
Angles 1 and 2 are "corresponding angles" and as such are equal. Thus, the measure of angle 2 is 60 degrees.