Your answer would be
A. Adding 7x to both sides of the equation
And in doing so, you'd start your first step to having "like terms" on "like sides"
by having all "x's" on the right side of your equivalent sign "="
Answer: 30 m ; (or, write as: "30 meters") .
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Explanation:
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Area of a trapezoid, "A" = (1/2) ( b₁ + b₂) h ;
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or, write as: A = ( b₁ + b₂) h / 2 ;
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in which: A = area;
b₁ = length of "base 1" (choose either one of the 2 (two bases);
b₂ = length of "base 2" (use the base that is remaining);
h = height of trapezoid;
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From the information given:
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A = 100 m² ;
h = 5 m
b₁ = 10 m
b₂ = x
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Find "x", which is: "b₂" ;
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A = ( b₁ + b₂) h / 2 ;
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Plug in our known values; and plug in "x" for "b₂" ; and solve for "x" ;
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100 m² = [(10m + x) (5m)] / 2 ; Solve for "x" ;
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(10m + x) (5m) = (2)* (100m²) ;
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(5m) (10m + x) = 200 m² ;
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Note: The distributive property of multiplication:
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a(b+c) = ab + ac ;
a(b−c) = ab <span>− ac ;
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We have: (5m) (10m + x) = 200 m² ;
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So: (5m) (10m + x) = (5m*10m) + (5m * x) ;
= 50m² + (5m)x ;
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→ 50m² + (5m)x = 200m² ;
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Divide the ENTIRE equation by "5m" ;
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→ { 50m² + (5m)x } / 5m = (200m² / 5m) ;
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→ 10m + x = 40m ;
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Now, subtract "10m" from EACH side of the equation; to isolate "x" on one side of the equation; and to solve for "x" ;
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→ 10m + x − 10m = 40m − 10m ;
to get:
→ x = 30 m ; which is our answer.
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Answer: 30 m ; (or, write as: "30 meters") .
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Y=2x+2
(y=mx+c
m-slope/gradient
c-y intercept)
Answer:
B:14
Step-by-step explanation:
Answer:
Is their an equation that can be used a reference to find the sum/difference/product/quotient to this equation?
Step-by-step explanation:
