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TEA [102]
3 years ago
14

Please help, I can't figure out how to set this up. There's supposed to be 2 equations.

Mathematics
1 answer:
Valentin [98]3 years ago
7 0

Hello from MrBillDoesMath

Answer:     + 4/7,   6 (13/14)

Discussion:

Let L be the larger of the two numbers and S the smaller of the two. Then the first sentence translates into this:


-6L - L = -4  =>

-7L = -4        =>          (divide both sides by -7)

L = - (-4)/(-7) = + 4/7       (*)


The second sentence translates to

2 (L + S) = 15  =>               (divide both sides by 2)

(L +S) = 15/2  =>                (subtract L from each side)

S = 15/2 - L = 15/2 - 4/7    (using (*) above)

So

S = 15*7/(2*7) - (4*2)/(7*2)       (combine over common denominator of 14)

S = 105/14 - 8/14 = 97/14 = 6 13/14

Something is crazy here!    While 2(L +S) = 15  -- check it out -- our larger number L ( =4/7) is actually smaller than the S number (6 13/14)!  Maybe I should have called them Y and X, respectively, as you suggested!

Thank you,

Mr. B


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Read 2 more answers
In the expansion (ax+by)^7, the coefficients of the first two terms are 128 and -224, respectively. Find the values of a and b
madam [21]

Answer:

a = 2, b = 3.5

Step-by-step explanation:

Expanding (ax+by)^7 using Binomial expansion, we have that:

(ax+by)^7 =

(ax)^7(by)^0 + (ax)^6(by)^1 + (ax)^5(by)^2 + (ax)^4(by)^3 + (ax)^3(by)^4 + (ax)^2(by)^5 + (ax)^1(by)^6 + (ax)^0(by)^7

= (a)^7(x)^7+ (a)^6(x)^6(b)(y) + (a)^5(x)^5(b)^2(y)^2 + (a)^4(x)^4(b)^3(y)^3 + (a)^3(x)^3(b)^4(y)^4 + (a)^2(x)^2(b)^5(y)^5 + (a)(x)(b)^6(y)^6 + (b)^7(y)^7\\\\\\= (a)^7(x)^7+ (a)^6(b)(x)^6(y) + (a)^5(b)^2(x)^5(y)^2 + (a)^4(b)^3(x)^4(y)^3 + (a)^3(b)^4(x)^3(y)^4 + (a)^2(b)^5(x)^2(y)^5 + (a)(b)^6(x)(y)^6 + (b)^7(y)^7

We have that the coefficients of the first two terms are 128 and -224.

For the first term:

=> a^7 = 128

=> a = \sqrt[7]{128}\\ \\\\a = 2

For the second term:

a^6b = -224

b = \frac{-224}{a^6}

b = \frac{-224}{2^6} \\\\\\b = \frac{-224}{64} \\\\\\b = 3.5

Therefore, a = 2, b = 3.5

5 0
3 years ago
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