Answer:
30 Cents
Step-by-step explanation:
1.
height= 3
length= 5
width= 4
The number of cubes would be the result of the multiplication of the side measures. Doing so, we have:
3 x 5 x 4 = 60
We need 60 blocks
2
height= 4
length= 6
width= 4
The number of cubes would be the result of the multiplication of the side measures. Doing so, we have:
4 x 6 x 4 = 96
We need 96 blocks
3
height= 2
length= 3
width= 4
The number of cubes would be the result of the multiplication of the side measures. Doing so, we have:
2 x 3 x 4 = 24
We need 24 blocks
4
height= 4
length= 8
width= 6
The number of cubes would be the result of the multiplication of the side measures. Doing so, we have:
4 x 8 x 6 = 192
We need 192 blocks
5
height= 2
length= 6
width= 4
The number of cubes would be the result of the multiplication of the side measures. Doing so, we have:
2 x 6 x 4 = 48
We need 48 blocks
6
height= 1
length= 5
width= 3
The number of cubes would be the result of the multiplication of the side measures. Doing so, we have:
1 x 5 x 3 = 15
We need 15 blocks
7
Answer:
If both computers are working together, it will take 24 minutes to do the job
Step-by-step explanation:
It is given that,
There are two computers.The slower computer can send all the company's email in 60 minutes.
The faster computer can complete the same job in 40 minutes
<u>To find the LCM of 40 and 60</u>
LCM (40, 60) = 120
<u>To find efficiency of 2 computers</u>
Let x be the efficiency of faster computer and y be the efficiency of faster computer
x = 120/40 = 3
y = 120/60 = 2
then, x + y = 3 + 2 = 5
Therefore efficiency of both the computer work together = x + y =5
<u>To find the time taken to work both the computer together</u>
time = 120/5 = 24 minutes
Answer:
sorry kailangan ko ng points
Answer: The square root of π has attracted attention for almost as long as π itself. When you’re an ancient Greek mathematician studying circles and squares and playing with straightedges and compasses, it’s natural to try to find a circle and a square that have the same area. If you start with the circle and try to find the square, that’s called squaring the circle. If your circle has radius r=1, then its area is πr2 = π, so a square with side-length s has the same area as your circle if s2 = π, that is, if s = sqrt(π). It’s well-known that squaring the circle is impossible in the sense that, if you use the classic Greek tools in the classic Greek manner, you can’t construct a square whose side-length is sqrt(π) (even though you can approximate it as closely as you like); see David Richeson’s new book listed in the References for lots more details about this. But what’s less well-known is that there are (at least!) two other places in mathematics where the square root of π crops up: an infinite product that on its surface makes no sense, and a calculus problem that you can use a surface to solve.
Step-by-step explanation: this is the same paragraph The square root of π has attracted attention for almost as long as π itself. When you’re an ancient Greek mathematician studying circles and squares and playing with straightedges and compasses, it’s natural to try to find a circle and a square that have the same area. If you start with the circle and try to find the square, that’s called squaring the circle. If your circle has radius r=1, then its area is πr2 = π, so a square with side-length s has the same area as your circle if s2 = π, that is, if s = sqrt(π). It’s well-known that squaring the circle is impossible in the sense that, if you use the classic Greek tools in the classic Greek manner, you can’t construct a square whose side-length is sqrt(π) (even though you can approximate it as closely as you like); see David Richeson’s new book listed in the References for lots more details about this. But what’s less well-known is that there are (at least!) two other places in mathematics where the square root of π crops up: an infinite product that on its surface makes no sense, and a calculus problem that you can use a surface to solve.
