The solution of
is ![\frac{1}{28} \text{ or } 0.0357](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B28%7D%20%5Ctext%7B%20or%20%7D%200.0357)
<em><u>Solution:</u></em>
Given that we have to find the solution of ![\frac{3}{4} - \frac{5}{7}](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B4%7D%20-%20%5Cfrac%7B5%7D%7B7%7D)
To solve the given sum, first make the denominators of both the fractions same
This can be done by taking L.C.M of both denominators
<em><u>Step 1:</u></em>
L.C.M of 4 and 7:
The prime factor of 4 = 2 x 2
The prime factor of 7 = 7
For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.
The new superset list is
2, 2, 7
Multiply these factors together to find the LCM
LCM = 2 x 2 x 7 = 28
Thus L.C.M of denominators is 28
<em><u>Step 2:</u></em>
Solution of ![\frac{3}{4} - \frac{5}{7}](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B4%7D%20-%20%5Cfrac%7B5%7D%7B7%7D)
Multiply the denominator by a number to get 28 and multiply that same number with numerator also
![\frac{3}{4} - \frac{5}{7} = \frac{3 \times 7}{4 \times 7} - \frac{5 \times 4}{7 \times 4}=\frac{21}{28} - \frac{20}{28}\\\\\frac{21}{28} - \frac{20}{28} = \frac{21-20}{28} = \frac{1}{28}\\\\\frac{1}{28} = 0.0357](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B4%7D%20-%20%5Cfrac%7B5%7D%7B7%7D%20%3D%20%5Cfrac%7B3%20%5Ctimes%207%7D%7B4%20%5Ctimes%207%7D%20-%20%5Cfrac%7B5%20%5Ctimes%204%7D%7B7%20%5Ctimes%204%7D%3D%5Cfrac%7B21%7D%7B28%7D%20-%20%5Cfrac%7B20%7D%7B28%7D%5C%5C%5C%5C%5Cfrac%7B21%7D%7B28%7D%20-%20%5Cfrac%7B20%7D%7B28%7D%20%3D%20%5Cfrac%7B21-20%7D%7B28%7D%20%3D%20%5Cfrac%7B1%7D%7B28%7D%5C%5C%5C%5C%5Cfrac%7B1%7D%7B28%7D%20%3D%200.0357)
Thus solution of
is ![\frac{1}{28} \text{ or } 0.0357](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B28%7D%20%5Ctext%7B%20or%20%7D%200.0357)
Answer:
1414 kL
Step-by-step explanation:
The volume of the donut-shaped moat is the product of its surface area and depth. The area is the product of its centerline length and its width.
<h3>Moat area</h3>
The diameter of the centerline of the moat is (50 m -5 m) = 45 m. The length of that centerline is ...
C = πd = π(45 m) = 45π m
The area is this length times the width of the moat:
moat area = (45π m)(5 m) = 225π m²
<h3>Moat volume</h3>
The volume is the product of the area and the depth of the moat:
V = Ah = (225π m²)(2 m) = 450π m³ ≈ 1413.72 m³
1 cubic meter is 1000 liters, 1 kiloliter.
The volume of the moat is about 1414 kL.
The mean is the average of a set of numbers.
To find the mean of this data, form a number set by gathering all the numbers.
We need to find the average weekly allowance. To do this, each number in the number set should be the different allowances, and their quantity is the number of students who earned that allowance.
In this case, there would be seven 0s, five 3s, seven 5s, three 6s, and two 8s.
The numbers are:
0, 0, 0, 0, 0, 0, 0, 3, 3, 3, 3, 3, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 8, 8
To find the mean of these numbers, add then together then divide by the total amount of numbers.
This means doing:
(0 + 0 + 0 + 0 + 0 + 0 + 0 + 3 + 3 + 3 + 3 + 3 + 5 + 5 + 5 + 5 + 5 + 5 + 5 + 6 + 6 + 6 + 8 + 8) / 7 + 5 + 7 + 3 + 2
An easier formula could be used by using multiplication.
This would be [7(0) + 5(3) + 7(5) + 3(6) + 2(8)] / 24
This is a lot easier to read!
Now to solve it.
7 • 0 = 0
5 • 3 = 15
7 • 5 = 35
3 • 6 = 18
2 • 8 = 16
0 + 15 + 35 + 18 + 16 = 84
84 / 24 = 3.5
The mean is 3.5, or $3.50
This means that the average weekly allowance amongst these students is $3.50.
Hope this helps!
Answer:
12
Step-by-step explanation:
5^2 + b^2 = 13^2
25 + b^2 = 169
169-25= 144
take the square root of 144
answer: 12
Answer:
13p
Step-by-step explanation:
Add 7 + 6.
=13
13 + p = 13p
13p is the simplified answer.