Length and width of rectangle is 15 inches and 6 inches respectively
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Solution:</u></h3>
Given that area of a rectangle is 90 square inch
Ratio of length to the width = 5: 2.
Need to determine length and width of rectangle.
As ratio of length to the width is 5 : 2
Lets assume length of rectangle = 5x inches and width of rectangle = 2x inches.
<em><u>The formula for area of rectangle is given as:</u></em>
Substituting the given value of area of rectangle and assumed value of length and width of rectangle we get:
On solving the above expression for x we get
Hence length and width of rectangle is 15 inches and 6 inches.
Answer:
Step-by-step explanation:
= 648 the length(12m), width(6m), and height(9m) that all i can give you for now because this got me losing brain sells
Answer:
B
Step-by-step explanation:
Answer:
Step-by-step explanation:
1) Eliminate parentheses:
0.1x +18.8 = -4 +2x
22.8 = 1.9x . . . . . . . . . add 4 - 0.1x
12 = x . . . . . . . . . . . . . divide by 1.9
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2) Eliminate parentheses:
-16 +4x = 0.8x +12.8
3.2x = 28.8 . . . . . . . . add 16 - 0.8x
x = 9 . . . . . . . . . . . . . .divide by 3.2
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<em>Comments on the solutions</em>
The expression we add in each case eliminates the constant on one side of the equation and the variable term on the other side. That leaves an equation of the form ...
variable term = constant
We choose to eliminate the smaller variable term (the one with the coefficient farthest to the left on the number line). Then the constant we eliminate is the on on the other side of the equation. This choice ensures that the remaining variable term has a positive coefficient, tending to reduce errors.
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You can work these problems by methods that eliminate fractions. Here, the fractions are decimal values, so are not that difficult to deal with. In any event, it is good to be able to work with numbers in any form: fractions, decimals, integers. It can save some steps.
