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2 years ago

Wat is the average between 1,000, 2300,1000,2600

Mathematics

1 answer:

earnstyle [38]2 years ago

6 0

Answer:

1180.2

Step-by-step explanation:

The average of 1,000,2300,1000,2600 is 1180.2

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The sum of 9/26 and -4/13 is ____ put the answer or else I WILL REPORT YOU

Colt1911 [192]

Answer:

1/26

Step-by-step explanation:

3 0

2 years ago

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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

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wlad13 [49]

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