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3 years ago

Determine the measures of the angles requested using the picture below.

Mathematics

1 answer:

lapo4ka [179]3 years ago

5 0

SHDJJEHEH shdhshshshshduvuvuvububuyb

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Step-by-step explanation:

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3 years ago

What is the LCM of 80

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so is the question right?

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Which expression is equivalent to (x Superscript 27 Baseline y) Superscript one-third?.

marin [14]

To solve the problem we must know the basic exponential properties.

<h3>What are the basic exponent properties?</h3>

${a^m} \cdot {a^n} = a^{(m+n)}$

$\dfrac{a^m}{a^n} = a^{(m-n)}$

$\sqrt[m]{a^n} = a^{\frac{n}{m}}$

$(a^m)^n = a^{m\times n}$

$(m\times n)^a = m^a\times n^a$

The expression can be written as $x^9\sqrt[3]{y}$ .

Given to us

$(x^{27}y)^\frac{1}{3}$

$(x^{27}y)^\frac{1}{3}$

Using the exponential property $(m\times n)^a = m^a\times n^a$ ,

$=(x^{27}y)^\frac{1}{3}\\\\=x^{\frac{27}{3}}\times y^\frac{1}{3}\\\\=x^9\times y^\frac{1}{3}$

Using the exponential property $\sqrt[m]{a^n} = a^{\frac{n}{m}}$ ,

$=x^9\times y^\frac{1}{3}\\\\=x^9\times \sqrt[3]{y}\\\\=x^9 \sqrt[3]{y}$

Hence, the expression can be written as $x^9\sqrt[3]{y}$ .

Learn more about Exponent properties:

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2 years ago

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mart [117]

Answer:

$\int\ {(3x^2-2x+1)\,(x^3-x^2+x)^7} \, dx = \frac{1}{8} (x^3-x^2+x)^8+C$

tep-by-step explanation:

In order to find the integral:

$\int\ {(3x^2-2x+1)\,(x^3-x^2+x)^7} \, dx$

we can do the following substitution:

Let's call

$u=(x^3-x^2+x)$

Then

$du = (3x^2-2x+1) dx$

which allows us to do convert the original integral into a much simpler one of easy solution:

$\int\ {(3x^2-2x+1)\,(x^3-x^2+x)^7} \, dx = \int\ {u^7 \, du = \frac{1}{8} \,u^8 +C$

Therefore, our integral written in terms of "x" would be:

$\int\ {(3x^2-2x+1)\,(x^3-x^2+x)^7} \, dx = \frac{1}{8} (x^3-x^2+x)^8+C$

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3 years ago

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asambeis [7]

Answer:

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Step-by-step explanation:

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2 years ago

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Determine the measures of the angles requested using the picture below.​

Determine the measures of the angles requested using the picture below.