Complete the explanation to determine whether each statement is always, sometimes, or never true. If quadrilateral PQRS is a rho
1 answer:
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Answer:
7 3/8 + (-4 1/2) ÷ (-5 2/3) = 8 23/136
Step-by-step explanation:
1) First I turned all the mix numbers into improper fractions:
7 3/8 ----> ( 7(8)+3/8) = 59/8, 4 1/2 ----> (4(2)+1/2) = 9/2, 5 2/3 ----> (5(3)+2/3) = 17/3
So now it should look like this: 59/8 + (-9/2)÷(-17/3)
2) Now our goal is to divide both of the improper fractions (-9/2)÷(-17/3),
- We first apply our fraction rule: -a/-b = a/b (when we have two negatives they cancel out each other and make a positive)
Our Case, From this:-9/2 ÷ -17/3 = To This: 9/2 ÷ 17/3
3) Now we can divide the fractions using this rule: a/b ÷ c/d = a times d / b times
Our Case, From This: 9/2 ÷ 17/3 To This: 9(3)/2(17) Which Gives Us: 27/34
(9 x 3 = 27, 2 x 17= 34)
So now it looks like this: 59/8 +27/34
4) Our look goal is to have the same denominator (which is the bottom part of the fraction) which are 8 and 34
To find it we find the LCM or Least Common Multiple of 8 and 34
(The LCM of a, b is the smallest positive number that is divisible by both a and b) which in this case a and b are 8 and 34
LCM is 136
5) We adjust our two fractions based on the LCM,
(Multiply each numerator ( top part of the fraction) by the same amount of needed to multiply its corresponding denominator to turn it to the LCM 136.
From This: 59/8 and 27/34 To This: 1003/136 and 108/36 ( 59(17)/8 (17) = 1003/136, 27(4)/34(4) = 108/306
6) Finally we can add the numerator (1003 and 108) together: 1003+108= 1111 and now we are left with 1111/136
Then we turn our improper fraction back into a mix number: 1111/138= 8 23/136
