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Anuta_ua [19.1K]
2 years ago
10

If u(x) = x5 – x4 + x2 and v(x) = –x2, which expression is equivalent to (StartFraction u Over v EndFraction) (x)?

Mathematics
1 answer:
Maslowich2 years ago
5 0

Based on the calculations, an expression which is equivalent to [u/v](x) is: C. -x³ + x² - 1.

<u>Given the following data:</u>

  • u(x) = x⁵ – x⁴ + x²
  • v(x) = -x²

<h3>What is an expression?</h3>

An expression is a mathematical equation which is used to show the relationship that exist between two or more numerical quantities or variables.

In this exercise, we would evaluate the given expressions by factorizing the function u(x) as follows:

u(x) = x⁵ – x⁴ + x²

u(x) = x²(x³ – x² + 1)

Rewriting the expressions as a fraction, we have:

\frac{u(x)}{v(x)} = \frac{x^2(x^3 - x^2 + 1)}{-x^2}\\\\\frac{u(x)}{v(x)} = -(x^3 - x^2 + 1)\\\\\frac{u(x)}{v(x)} = -x^3 + x^2 - 1

Therefore, u(x)/v(x) = -x³ + x² - 1.

Read more on expressions here: brainly.com/question/12189823

#SPJ1

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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]

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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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