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denis23 [38]
3 years ago
13

Select the correct answer how many solutions for x dose the following equation have 2(x+4)-1=2x+7

Mathematics
1 answer:
gavmur [86]3 years ago
5 0

Answer:

every number is correct.

Step-by-step explanation:

2 (x + 4) - 1 = 2x + 7

2x + 8 - 1 = 2x + 7

2x + 7 = 2x + 7

     - 7          - 7

2x = 2x

/2     /2

x = x

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X^8+8x^7+15x^6−9x^5−72x^4−135x^3+8x^2+64x+120
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(x+3)(x+5)<em>(x-2)(x-1)</em><u>(x^2+2x+4)(x^2+x+1)</u> where the given one is bolded the 2 binomials is italicized  and the 2 trinomials is underlined

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3 years ago
What is G equivalent to in 21.8-G=9.5???
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G is equal 12.3 in this equation
7 0
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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
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