If we let p and t be the masses of the paper and textbook, respectively, the equations that would best represent the given in this item are:
(1) 20p + 9t = 44.4
(2) (20 + 5)p + (9 + 1)t = 51
The values of p and t from the equation are 0.6 and 3.6, respectively. Thus, each paperback weighs 0.6 pounds and each textbook weighs 3.6 pounds.
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Answer:
C) 7
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Work Shown:
Use the slope formula
m = (y2-y1)/(x2-x1)
Plug in the given slope we want m = -5/3 and also the coordinates of the points. Then isolate r
m = (y2-y1)/(x2-x1)
-5/3 = (2-r)/(r-4)
-5(r-4) = 3(2-r) .... cross multiplying
-5r+20 = 6-3r
-5r+20+5r = 6-3r+5r .... adding 5 to both sides
20 = 6+2r
20-6 = 6+2r-6 ....subtracting 6 from both sides
14 = 2r
2r = 14
2r/2 = 14/2 .... dividing both sides by 2
r = 7
The slope of the line through (4,7) and (7,2) should be -5/3, let's check that
m = (y2-y1)/(x2-x1)
m = (2-7)/(7-4)
m = -5/3
The answer is confirmed
Answer:
Numerator = 2(b^2+a^2) or equivalently 2b^2+2a^2
Denominator = (b+a)^2*(b-a), or equivalently b^3+ab^2-a^2b0-a^3
Step-by-step explanation:
Let
S = 2b/(b+a)^2 + 2a/(b^2-a^2) factor denominator
= 2b/(b+a)^2 + 2a/((b+a)(b-a)) factor denominators
= 1/(b+a) ( 2b/(b+a) + 2a/(b-a)) find common denominator
= 1/(b+a) ((2b*(b-a) + 2a*(b+a))/((b+a)(b-a)) expand
= 1/(b+a)(2b^2-2ab+2ab+2a^2)/((b+a)(b-a)) simplify & factor
= 2/(b+a)(b^2+a^2)/((b+a)(b-a)) simplify & rearrange
= 2(b^2+a^2)/((b+a)^2(b-a))
Numerator = 2(b^2+a^2) or equivalently 2b^2+2a^2
Denominator = (b+a)^2*(b-a), or equivalently b^3+ab^2-a^2b0-a^3
Answer:
3/5
Step-by-step explanation:
