A few days ago
Anonymous

Any math geniuses out there??? a challenge calls!!!?

if a,b,c, and d are non-zero integers and 2a=3b, b=1/2c, and 3c=d, what is the value of d/a?

please give me a detailed answer…. best one gets 10pts.!!

Top 7 Answers
A few days ago
hayharbr

Favorite Answer

OK. Suppose just to make things come out even, that c = 12. Then d would be 36 since 3c = d. b would be 6 since it’s half of c. 3b would be 18, equalling 2a, so a would be 9.

That makes d/a = 36/9 or 4/1.

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A few days ago
chevy_vg
Given, 2a = 3b, b = 1/2 c and 3c = d,

what is the value of d/a?

Use the substitution method.

2a = 3b

2a = 3 (1/2c) since b = 1/2c

2a = 3/2 c

2a = d/2 since 2c = d

4 = d/a cross multiplication

d/a = 4

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A few days ago
Drew W
d/a = 4

2a=3b b=1/2c 3c=d so if you start substituting values. sub b=1/2c into 2a=3b

2a = 3/2c c actually equals d/3 so sub this in to 2a = 3/2* d/3

and you get 2a = 3d/6 at the point simplify and you get 12a = 3d.

solve for d/a by dividing both sides by a .. this gives 12 = 3d/a

divide 3 into 12 and d/a = 4

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A few days ago
Mysterious
d = 3c (given)

2a = 3b (given)=> a=3b/2

d\a = 3c \ 3b * 2

= 2c\b

b=1 /2 c => 2b = c ; 2c = 4b

substitute this in the previous equation

4b\b = 3b

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A few days ago
LeSsBrAiN
Given: 2a = 3b

b = 1/2c

3c = d

Required: d/a

solution:

a=3b/2 where b = 0.5c

a=1.5c/2

d/a = 3c/(1.5c/2)

d/a = 6/1.5

d/a = 4

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4 years ago
?
you have one perfect-angled triangle it relatively is broken up into 2 perfect-angled triangles, so for you to use Pythagoras thrice: 8^2 + y^2 = z^2 (14 – 8)^2 + y^2 = x^2 x^2 + z^2 = 14^2 From the 1st equation, y^2 = z^2 – 8^2 From the third equation, x^2 = 14^2 – z^2 replace those 2 into the 2d equation: (14 – 8)^2 + (z^2 – 8^2) = 14^2 – z^2 6^2 + z^2 – sixty 4 = 196 – z^2 36 + z^2 – sixty 4 = 196 – z^2 z^2 + z^2 = 196 – 36 + sixty 4 2z^2 = 224 z^2 = 224/2 z^2 = 112 because z^2 = 112, replace that into the 1st equation: 8^2 + y^2 = 112 sixty 4 + y^2 = 112 y^2 = 112 – sixty 4 y^2 = 40 8 because z^2 = 112, replace that into the third equation: x^2 + 112 = 14^2 x^2 + 112 = 196 x^2 = 196 – 112 x^2 = eighty 4 we now have: x^2 = eighty 4 x = sqrt(eighty 4) x = sqrt(4 * 21) x = sqrt(4) * sqrt(21) x = 2 * sqrt(21) y^2 = 40 8 y = sqrt(40 8) y = sqrt(sixteen * 3) y = sqrt(sixteen) * sqrt(3) y = 4 * sqrt(3) z^2 = 112 z = sqrt(112) z = sqrt(sixteen * 7) z = sqrt(sixteen) * sqrt(7) z = 4 * sqrt(7)
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A few days ago
reality
a=3b/2, b=c/2, c=d/3, d=3c

d/a=6c/3b=2c/b=4

d/a=4

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