How did you compute sums of dollar amounts that were not whole numbers? For example, how did you compute the sum of$5.89 and$1.4
5? Use this example to explain your strategy. 2.
2 answers:
Answer:
$7.34
Step-by-step explanation:
To compute sum of dollars that are not whole numbers. Using the sum of$5.89 and$1.45 as an illustration :
$5.89 + $1.45
Taking the whole numbers first:
$5 + $1 = $6
Take the sum of the decimals :
$0.89 + $0.45 = $1.34
Sum initial whole + whole of sum of decimal
$6 + $1 = $7
Remaining decimal : $1.34 - $1 = $0.34
$7 + $0.34 = $7.34
Answer:
$7.34
Step-by-step explanation:
To compute sum of dollars that are not whole numbers. Using the sum of$5.89 and$1.45 as an illustration :
$5.89 + $1.45
Taking the whole numbers first:
$5 + $1 = $6
Take the sum of the decimals :
$0.89 + $0.45 = $1.34
Sum initial whole + whole of sum of decimal
$6 + $1 = $7
Remaining decimal : $1.34 - $1 = $0.34
$7 + $0.34 = $7.34
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Answer:
8-2=6 the answer is 6 because there are 4 cps in 1 quart
Step-by-step explanation:
5x+3(x-2)=90
5x+3x-6=90
8x-6=90
+6 +6
8x=96
\8 \8
X=12
Answer:
Length = 9 inches
Step-by-step explanation:
The perimeter of a rectangle is:
perimeter = 2(length + width)
24 = 2(a+b)
a = b + 6
a = length
b = width
then:
24 = 2((b+6)+b)
24/2 = b+6+b
12 = 2b + 6
12-6 = 2b
6 = 2b
b = 6/2
b = 3 inches
a = b+6
a = 3+6
a = 9 inches
Check:
24 = 2(9+3)
24 = 2*12
412 + 325 = R
R = 737
I hope that is the answer you were looking for!
