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solmaris [256]
1 year ago
11

Which of the following is a step in simplifying the expression x multiplied by y to the power of 4 over x to the power of negati

ve 5 multiplied by y to the power of 5, the whole to the power of negative 3.? (5 points)
Group of answer choices

x to the power of negative 3 multiplied by y to the power of negative 12, the whole over x to the power of 15 multiplied by y to the power of negative 15.

x to the power of negative 3 multiplied by y, the whole over x to the power of negative 8 multiplied by y to the power of 2.

x to the power of negative 3 multiplied by y, the whole over x to the power of negative 5 multiplied by y to the power of 5.

x to the power of negative 3 multiplied by y to the power of negative 12, the whole over x to the power of negative 5 multiplied by y to the power of 5.

Flag this Question
Question 6
Mathematics
1 answer:
vodomira [7]1 year ago
4 0

Answer:

x to the power of negative 3 multiplied by y to the power of negative 12, the whole over x to the power of 15 multiplied by y to the power of negative 15.

Step-by-step explanation:

Given:

(\frac{xy^4}{x^{-5}y^5} )^{-3}

We need to simplify the equation.

As while solving these kind of problems, keep in mind the following Law on Indices:

1. (a^m)^n=a^{mn}

Applying the same we get;

\frac{x^{-3}(y^4)^{-3}}{(x^{-5})^{-3}(y^5)^{-3}}\\\\\frac{x^{-3}y^{4\times-3}}{x^{-5\times-3}y^{5\times-3}} \\\\\frac{x^{-3}y^{-12}}{x^{15}y^{-15}}

Final Answer:

x to the power of negative 3 multiplied by y to the power of negative 12, the whole over x to the power of 15 multiplied by y to the power of negative 15.

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The quadratic equations and their solutions are;

9 ± √33 /4 = 2x² - 9x + 6.

4 ± √6 /2 = 2x² - 8x + 5.

9 ± √89 /4 = 2x² - 9x - 1.

4 ± √22 /2 = 2x² - 8x - 3.

Explanation:

Any quadratic equation of the form, ax² + bx + c = 0 can be solved using the formula x = -b ± √b² - 4ac / 2a. Here a, b, and c are the coefficients of the x², x, and the numeric term respectively.

We have to solve all of the five equations to be able to match the equations with their solutions.

2x² - 8x + 5, here a = 2, b = -8, c = 5.                                                  x = -b ± √b² - 4ac / 2a = -(-8) ± √(-8)² - 4(2)(5) / 2(2) = 8 ± √64 - 40/4.     24 can also be written as 4 × 6 and √4 = 2. So                                                                                     x = 8 ± 2√6 / 2×2= 4±√6/2.

2x² - 10x + 3, here a = 2, b = -10, c = 3.                                                   x =-b ± √b² - 4ac / 2a =-(-10) ± √(-10)² - 4(2)(3) / 2(4) = 10 ± √100 + 24/4. 124 can also be written as 4 × 31 and √4 = 2. So                                                                              x = 10 ± 2√31 / 2×2 = 5 ± √31 /2.

2x² - 8x - 3, here a = 2, b = -8, c = -3.                                                    x = -b ± √b² - 4ac / 2a = -(-8) ± √(-8)² - 4(2)(-3) / 2(2) = 8 ± √64 + 24/4.     88 can also be written as 4 × 22 and √4 = 2. So                                                                             x = 8 ± 2√22 / 2×2 = 4± √22/2.

2x² - 9x - 1, here a = 2, b = -9, c = -1.                                                     x = -b ± √b² - 4ac / 2a = -(-9) ± √(-9)² - 4(2)(-1) / 2(2) = 9 ± √81 + 8/4.                                          x = 9 ± √89 / 4.

2x² - 9x + 6, here a = 2, b = -9, c = 6.                                                    x = -b ± √b² - 4ac / 2a = -(-9) ± √(-9)² - 4(2)(6) / 2(2) = 9 ± √81 - 48/4.                                                                             x = 9 ± √33 / 4

To match we solve the monomials.

1. -15u^3 + 5u^3

Adding

-15u^3 + 5u^3=-10u^3

2.  10u^3 +(-5u^3)

Adding

10u^3-5u^3=5u^3

3. 10u^3 + 5u^3

Adding

10u^3 + 5u^3=15u^3

4.  5u^3+ (-10u^3)

Adding

5u^3-10u^3 =-5u^3

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