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3 years ago

What's the function rule, I don't understand how to write a function rule using the ordered pairs in this table

Mathematics

1 answer:

pickupchik [31]3 years ago

4 0

The rule is plus four because every time you take your x value and add four it equals your y value.

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Element X is a radioactive isotope such that every 25 years, its mass decreases by

luda_lava [24]

Answer: 26

Step-by-step explanation:

6 0

3 years ago

If the product of 0x0 the same as the sum of 0+0? Explain

umka21 [38]

They are both the same because anything times 0 equal 0 and 0+0 equals 0.

5 0

3 years ago

Read 2 more answers

Laurie buys four bags of candy with 10 pieces of candy in each bag. Her father then gives her 5 more pieces of candy. Laurie wan

Alborosie

Your Question:
Laurie buys four bags of candy with 10 pieces of candy in each bag. Her father then gives her 5 more pieces of candy. Laurie wants to give as much of her candy as she can to each of her six friends, and she wants to make sure they each get an equal number of pieces. How many pieces will be left over, if any, after Laurie gives her friends the candy?
My Answer:
If Laurie has four bags with ten pieces in each bag, and if her father gives her five more pieces, she would have forty-five pieces and if she has six friends and wants to share it equally with six friends. Each friend would get 7 pieces.
My Work:
45 ÷ 6 = 7.5
Hope I helped
♥ James.

7 0

3 years ago

Read 2 more answers

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.