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Anni [7]
3 years ago
11

Evaluate. 15 + 102 • [12 – (12 – 3 • 2)] + 6 and you can put 3 numbers only 2

Mathematics
2 answers:
mylen [45]3 years ago
6 0
Okay, lets do it15+102*[12-(12-3*2)]+6\\
15+102*[12-6]+6\\
12+102*6+6\\
15+612+6\\
=633

Dennis_Churaev [7]3 years ago
5 0
Use pemdas so you could do that in order
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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

5 0
3 years ago
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