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Gennadij [26K]
3 years ago
7

Find the value of (x+5) (x+2)+(x-3) (x-4) in its simplest form. What is the numerical value when x = -6?​

Mathematics
2 answers:
Usimov [2.4K]3 years ago
4 0

Answer:

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Alex777 [14]3 years ago
3 0

Answer:

( - x + 5)(x + 2) + (x - 3)(x - 4)

where as; x = - 6

<u>Solution</u>

<u>(x + 5)(x + 2) + (x - 3)( x - 4)</u>

<u>( - 6 + 5)( - 6 + 2) + ( - 6 - 3)( - 6 - 4)</u>

<u>(- 1)( - 4) + ( - 9)( - 10)</u>

<u>4 + 90</u>

<u>94</u>

The answer is 94.

<h3>Hope it's correct!</h3><h3><em>Please mark me as brainliest</em><em>☺</em><em>!</em><em>!</em></h3>

<em>Thanks</em><em>☺</em><em>!</em><em>!</em><em>!</em>

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Answer:

Solving the expression \frac{y^2z^{\frac{1}{4}} }{(z^{\frac{1}{2}}.xy^{\frac{3}{2}})^3} we get \mathbf{\frac{1}{y^{\frac{5}{2} }x^3z^{\frac{5}{4} }}}

Step-by-step explanation:

We need to solve the expression:

\frac{y^2z^{\frac{1}{4}} }{(z^{\frac{1}{2}}.xy^{\frac{3}{2}})^3}

We know the exponent rule: (a^n)^m = a^{nm}

\frac{y^2z^{\frac{1}{4}} }{(z^{\frac{1}{2}}.xy^{\frac{3}{2}})^3}\\\\=\frac{y^2z^{\frac{1}{4}} }{z^{\frac{3}{2}}.x^3y^{\frac{9}{2} }}

Now, another exponent rule says that: \frac{a^m}{a^n}=a^{m-n}

=\frac{y^{2-\frac{9}{2}} z^{\frac{1}{4}-\frac{3}{2} } }{x^3}\\=\frac{y^{\frac{4-9}{2}} z^{\frac{1-3*2}{4} } }{x^3}\\=\frac{y^{\frac{-5}{2}}z^{\frac{-5}{4} } }{x^3}

We also know that: a^{-m}=\frac{1}{a^m}

=\frac{1}{y^{\frac{5}{2} }x^3z^{\frac{5}{4} }}

So, solving the expression \frac{y^2z^{\frac{1}{4}} }{(z^{\frac{1}{2}}.xy^{\frac{3}{2}})^3} we get \mathbf{\frac{1}{y^{\frac{5}{2} }x^3z^{\frac{5}{4} }}}

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