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1 year ago

Identify the like terms in the list.

Mathematics

1 answer:

MrRa [10]1 year ago

8 0

Answer:

17x and 3x

Step-by-step explanation:

like terms are terms that have the same variables and powers. The coefficients do not need to match. Unlike terms are two or more terms that are not like terms, i.e. they do not have the same variables or powers. The order of the variables does not matter unless there is a power.

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What is the product of 5 + 2i and 7 + 3i?

ale4655 [162]

The answer is (4) because you multiply the like terms in the equation. 5*7=35 and 3i*2i=6i.

Answer is (4)

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3 years ago

01.01) Rewrite the expression with rational exponents as a radical expression. 7 times x to the two thirds power cube root of th

jeka94

Answer:

$(7x)^{\frac{2}{3} = (\sqrt[3]{7x})^2$

Step-by-step explanation:

Given the expression $(7x)^{\frac{2}{3}$ , to express this as a radical expressions, we'd apply the rule/law of indices that deals with converting expressions that has rational exponents into radical expressions.

The rule of indices to apply is: $b^{\frac{m}{n}} = (\sqrt[n]{b})^m$

To apply this to the expression, $(7x)^{\frac{2}{3}$ , the denominator of the fraction of the exponent would determine the root, that is, cube root in this case. The numerator of the exponent would then determine the exponent of the radical expressions.

Thus:

$(7x)^{\frac{2}{3} = (\sqrt[3]{7x})^2$

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3 years ago

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vfiekz [6]

Step-by-step explanation:

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2 years ago

Write the given expression as an algebraic expression in x. tan(2 cos^-1(x))

miv72 [106K]

$\bf tan\left[ 2cos^{-1}(x) \right]\implies tan\left[ 2cos^{-1}\left( \frac{x}{1} \right) \right]
\\\\\\
\textit{if we say }cos^{-1}\left( \frac{x}{1} \right)=\theta\textit{ that means }tan\left[ 2cos^{-1}\left( \frac{x}{1} \right) \right]\iff tan(2\theta)\\\\
-----------------------------\\\\
cos^{-1}\left( \frac{x}{1} \right)=\theta\implies cos(\theta)=\cfrac{x}{1}\cfrac{\leftarrow adjacent=a}{\leftarrow hypotenuse=c}\\\\
-----------------------------\\\\
$

$\bf \textit{again, using the pythagorean theorem to get the opposite side}
\\\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b\implies \pm\sqrt{1^2-x^2}=b
\\\\\\
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-----------------------------\\\\
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$\bf tan(2\theta)=\cfrac{\frac{\pm2\sqrt{1-x^2}}{x}}{1-\frac{1-x}{x}}\implies 
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tan(2\theta)=\cfrac{\pm 2\sqrt{1-x^2}}{2x-1}$

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