together = add
add the fractions
3 7/10 + 5 9/10
add the whole numbers first
3+5=8
add the fractions
7/10+9/10= 16/10
8 16/10
reduce 16/10 divide by 2
16/2= 8
10/2= 5
8 8/5
whole number changes to 9
8-5=3
denominator stays the same
9 3/5
Answer:
9 3/5
Answer:
It is irrational because it can not be represented as a fraction of two integers
Step-by-step explanation:
Given
![B_1 = 3.6](https://tex.z-dn.net/?f=B_1%20%3D%203.6)
![B_2 = 12\frac{1}{3}](https://tex.z-dn.net/?f=B_2%20%3D%2012%5Cfrac%7B1%7D%7B3%7D)
![H = \sqrt[3]5](https://tex.z-dn.net/?f=H%20%3D%20%5Csqrt%5B3%5D5)
Required
Why is the area irrational?
First, we need to calculate the area
![Area = \frac{1}{2}(B_1 + B_2) * H](https://tex.z-dn.net/?f=Area%20%3D%20%5Cfrac%7B1%7D%7B2%7D%28B_1%20%2B%20B_2%29%20%2A%20H)
![Area = \frac{1}{2}(3.6 + 12\frac{1}{3}) * \sqrt[3]5](https://tex.z-dn.net/?f=Area%20%3D%20%5Cfrac%7B1%7D%7B2%7D%283.6%20%2B%2012%5Cfrac%7B1%7D%7B3%7D%29%20%2A%20%5Csqrt%5B3%5D5)
![Area = \frac{1}{2}(3.6 + \frac{37}{3}) * 17100](https://tex.z-dn.net/?f=Area%20%3D%20%5Cfrac%7B1%7D%7B2%7D%283.6%20%2B%20%5Cfrac%7B37%7D%7B3%7D%29%20%2A%2017100)
![Area = \frac{1}{2}(\frac{10.8+37}{3}) * 1.710](https://tex.z-dn.net/?f=Area%20%3D%20%5Cfrac%7B1%7D%7B2%7D%28%5Cfrac%7B10.8%2B37%7D%7B3%7D%29%20%2A%201.710)
![Area = \frac{1}{2}(\frac{47.8}{3}) * 1.7100](https://tex.z-dn.net/?f=Area%20%3D%20%5Cfrac%7B1%7D%7B2%7D%28%5Cfrac%7B47.8%7D%7B3%7D%29%20%2A%201.7100)
![Area = \frac{47.8}{6} * 1.7100](https://tex.z-dn.net/?f=Area%20%3D%20%5Cfrac%7B47.8%7D%7B6%7D%20%2A%201.7100)
![Area = \frac{47.8* 1.7100}{6}](https://tex.z-dn.net/?f=Area%20%3D%20%5Cfrac%7B47.8%2A%201.7100%7D%7B6%7D)
![Area = \frac{81.738}{6}](https://tex.z-dn.net/?f=Area%20%3D%20%5Cfrac%7B81.738%7D%7B6%7D)
![Area = 13.623](https://tex.z-dn.net/?f=Area%20%3D%2013.623)
<em>It is irrational because it can not be represented as a fraction of two integers</em>
Hello!
Let's put this into a system of equations and have a represent the apples and o represent the oranges.
o=2a+2
0.5o-0.5a=4
Since we can already so what o equals (2a+2) we will substitute it into the second equation and solve for a.
0.5(2a+2)-0.5a=4
1a+1-0.5a=4
1.5a+1=4
1.5a=3
a=2
Now that we know the value of a will put a into the first equation to find o.
o=2(2)+2
o=4+2
o=6
There were 6 oranges and 2 apples.
I hope this helps!
Second one is correct, that is an identity.
Third one is a contradiction:
6x - 15 = 3 * (2x - 6) ------> 6x - 15 = 6x - 18
Fourth one:
5x - 3 * (x - 4) = 2 * (x - 6) ---> 5x - 3x + 12 = 2x - 12 ---> 2x + 12 = 2x -12
So this is a contradiction as well.