Change 5% to a decimal which is 0.05
Then multiply 0.05 by 220 which is 11.
Answer: Meaghan is right.
Step-by-step explanation:
For a number like:
123.45
the tens place would be the "2" (second number at the left of the decimal point)
The ones place would be the "3" (first number at the left of the decimal point)
Let's suppose that Lily is correct.
Then the quotient of:
43.61/7
Will be a number with two digits in the left of the decimal point.
The smallest number that meets this condition, is the number 10.
Then let's see:
7*10 = 70
then:
70/7 = 10
and:
43.61 < 70
(70 is larger than 43.61)
then:
43.61/7 < 70/7 = 10
Then:
(43.61/7) < 10
This means that our quotient is smaller than 10, then the first digit of the quotient can not be on the tens place.
Then Meaghan is the correct one.
We also could perform the quotient to find:
43.61/7 = 6.23
So yes, Meaghan is correct.
Answer:
49
Step-by-step explanation:
the line joined by midpoints of sides is called midsection
which is parallel to the third side and is half its length
so (5x - 26) * 2 = 9x - 37
10x - 52 = 9x - 37
x = 52 - 37
x = 15
so PQ is 5x - 26
= 5*15 - 26
= 75 - 26
= 49
Answer:
![a^3 + 1 = 0](https://tex.z-dn.net/?f=a%5E3%20%2B%201%20%3D%200)
Step-by-step explanation:
We start with the equation:
![a + \frac{1}{a} = 1](https://tex.z-dn.net/?f=a%20%2B%20%5Cfrac%7B1%7D%7Ba%7D%20%3D%201)
We want to find the value of:
![a^3 + 1 =](https://tex.z-dn.net/?f=a%5E3%20%2B%201%20%3D)
We can start with our previous equation and multiply both sides by a:
![(a + \frac{1}{a})*a = 1*a\\a^2 + 1 = a](https://tex.z-dn.net/?f=%28a%20%2B%20%5Cfrac%7B1%7D%7Ba%7D%29%2Aa%20%3D%201%2Aa%5C%5Ca%5E2%20%2B%201%20%3D%20a)
Now we can rewrite our initial expression as:
![a = 1 - \frac{1}{a}](https://tex.z-dn.net/?f=a%20%3D%201%20-%20%5Cfrac%7B1%7D%7Ba%7D)
Replacing that in the right side, we get:
![a^2 + 1 = a = 1 - \frac{1}{a}](https://tex.z-dn.net/?f=a%5E2%20%2B%201%20%3D%20a%20%3D%201%20-%20%5Cfrac%7B1%7D%7Ba%7D)
Now again, let's multiply both sides by a
![a*(a^2 + 1) = a*(1 - \frac{1}{a} )\\a^3 + a = a - a/a\\a^3 + a = a - 1\\a^3 = -1\\a^3 + 1 = 0](https://tex.z-dn.net/?f=a%2A%28a%5E2%20%2B%201%29%20%3D%20a%2A%281%20-%20%5Cfrac%7B1%7D%7Ba%7D%20%29%5C%5Ca%5E3%20%2B%20a%20%3D%20a%20-%20a%2Fa%5C%5Ca%5E3%20%2B%20a%20%3D%20a%20-%201%5C%5Ca%5E3%20%3D%20-1%5C%5Ca%5E3%20%2B%201%20%3D%200)
So we can conclude that:
![a^3 + 1 = 0](https://tex.z-dn.net/?f=a%5E3%20%2B%201%20%3D%200)