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2 years ago

What's the ordered pair for point H in the graph above?

Mathematics

1 answer:

Bumek [7]2 years ago

6 0

Answer:

C) (5,0)

Step-by-step explanation:

1. The correct answer is C) because when we move five units to right from the origin, and zero units up, we end up at point H!

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Simplify this please

Ugo [173]

Answer:

$\frac{12q^{\frac{7}{3}}}{p^{3}}$

Step-by-step explanation:

Here are some rules you need to simplify this expression:

Distribute exponents: When you raise an exponent to another exponent, you multiply the exponents together. This includes exponents that are fractions. $(a^{x})^{n} = a^{xn}$

Negative exponent rule: When an exponent is negative, you can make it positive by making the base a fraction. When the number is apart of a bigger fraction, you can move it to the other side (top/bottom). $a^{-x} = \frac{1}{a^{x}}$ , and to help with this question: $\frac{a^{-x}b}{1} = \frac{b}{a^{x}}$ .

Multiplying exponents with same base: When exponential numbers have the same base, you can combine them by adding their exponents together. $(a^{x})(a^{y}) = a^{x+y}$

Dividing exponents with same base: When exponential numbers have the same base, you can combine them by subtracting the exponents. $\frac{a^{x}}{a^{y}} = a^{x-y}$

Fractional exponents as a radical: When a number has an exponent that is a fraction, the numerator can remain the exponent, and the denominator becomes the index (example, index here ∛ is 3). $a^{\frac{m}{n}} = \sqrt[n]{a^{m}} = (\sqrt[n]{a})^{m}$

$\frac{(8p^{-6} q^{3})^{2/3}}{(27p^{3}q)^{-1/3}}$ Distribute exponent

$=\frac{8^{(2/3)}p^{(-6*2/3)}q^{(3*2/3)}}{27^{(-1/3)}p^{(3*-1/3)}q^{(-1/3)}}$ Simplify each exponent by multiplying

$=\frac{8^{(2/3)}p^{(-4)}q^{(2)}}{27^{(-1/3)}p^{(-1)}q^{(-1/3)}}$ Negative exponent rule

$=\frac{8^{(2/3)}q^{(2)}27^{(1/3)}p^{(1)}q^{(1/3)}}{p^{(4)}}$ Combine the like terms in the numerator with the base "q"

$=\frac{8^{(2/3)}27^{(1/3)}p^{(1)}q^{(2)}q^{(1/3)}}{p^{(4)}}$ Rearranged for you to see the like terms

$=\frac{8^{(2/3)}27^{(1/3)}p^{(1)}q^{(2)+(1/3)}}{p^{(4)}}$ Multiplying exponents with same base

$=\frac{8^{(2/3)}27^{(1/3)}p^{(1)}q^{(7/3)}}{p^{(4)}}$ 2 + 1/3 = 7/3

$=\frac{\sqrt[3]{8^{2}}\sqrt[3]{27}p\sqrt[3]{q^{7}}}{p^{4}}$ Fractional exponents as radical form

$=\frac{(\sqrt[3]{64})(3)(p)(q^{\frac{7}{3}})}{p^{4}}$ Simplified cubes. Wrote brackets to lessen confusion. Notice the radical of a variable can't be simplified.

$=\frac{(4)(3)(p)(q^{\frac{7}{3}})}{p^{4}}$ Multiply 4 and 3

$=\frac{12pq^{\frac{7}{3}}}{p^{4}}$ Dividing exponents with same base

$=12p^{(1-4)}q^{\frac{7}{3}}$ Subtract the exponent of 'p'

$=12p^{(-3)}q^{\frac{7}{3}}$ Negative exponent rule

$=\frac{12q^{\frac{7}{3}}}{p^{3}}$ Final answer

Here is a version in pen if the steps are hard to see.

5 0

3 years ago

Which of the following represents the sum of the series? 6 8 10 12 14 16 18 20 22 24 26

stira [4]

The sum of the arithmetic series 6 8 10 12 14 16 18 20 22 24 26 is 176

<h3>How to determine the sum of the series?</h3>

The series is given as:

6 8 10 12 14 16 18 20 22 24 26

The above series is an arithmetic series with the following parameters:

First term, a = 6

Last term, L = 26

Number of terms, n = 11

The sum of the series is calculated using:

$S_n= \frac{n}{2} *(a + L)$

This gives

$S_n= \frac{11}{2} *(6 + 26)$

Evaluate

$S_{11} = 176$

Hence, the sum of the arithmetic series 6 8 10 12 14 16 18 20 22 24 26 is 176

Read more about arithmetic series at:

brainly.com/question/6561461

#SPJ4

7 0

2 years ago

Which statement is correct about y = cos^–1 x?

Maurinko [17]

Answer:

A) If the domain of y=cos(x) is restricted to [0, π], y=cos^-1(x) is a function.

Step-by-step explanation:

In order for the inverse function to be a function, the original must pass the horizontal line test: a horizontal line must intersect the function in only one place.

As you can see from the attached graph, restricting the cosine function to the domain [0, π] allows it to pass the horizontal line test, so its inverse will be a function.

Restricting the domain to [-π/2, π/2] does not limit cos(x) to something that will pass the horizontal line test.