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Juli2301 [7.4K]
3 years ago
9

Solve for x:

} = 9" alt="(x - 5)^{2} = 9" align="absmiddle" class="latex-formula">
​
Mathematics
2 answers:
Valentin [98]3 years ago
5 0
X-5=3 as the square root of 9 is 3
X=8 plus 5
Hope this helps
Sedaia [141]3 years ago
3 0

Answer:

Step-by-step explanation:

(x-5)^2=9

(x-5)(x-5)=9

x^2-10x+25=9

x^2-10x+16=0

(x-8)(x-2)

x=8,2

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What is the least common denominator of the four fractions listed below 20 7/10, 20 3/4, 18 9/10, 20 18/25
Feliz [49]
So what you do is see all the denomenators and factor them (denomenaotrs are the bottom numbres)

10,4,10,25
10=2 times 5
4=2 times 2
10=2 times 5
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we need to include all of them so we need at leas two 2's, and two 5's
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Use the rules of exponents to simplify the expressions. Match the expression with its equivalent value.
Lelechka [254]

Answer:

1) \frac{(-2)^{-5}}{(-2)^{-10}}=-32

2) 2^{-1}.2^{-4} = \frac{1}{32}

3) (-\frac{1}{2} )^3.(-\frac{1}{2} )^2=-\frac{1}{32}

4) \frac{2}{2^{-4}} = 32

Step-by-step explanation:

1) \frac{(-2)^{-5}}{(-2)^{-10}}

Solving using exponent rule: a^{-m}=\frac{1}{a^m}

\frac{(-2)^{-5}}{(-2)^{-10}}\\=(-2)^{-5+10}\\=(-2)^{5}\\=-32

So, \frac{(-2)^{-5}}{(-2)^{-10}}=-32

2) 2^{-1}.2^{-4}

Using the exponent rule: a^m.a^n=a^{m+n}

We have:

2^{-1}.2^{-4}\\=2^{-1-4}\\=2^{-5}

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

Using this rule:

2^{-5}\\=\frac{1}{2^5}\\=\frac{1}{32}

So, 2^{-1}.2^{-4} = \frac{1}{32}

3) (-\frac{1}{2} )^3.(-\frac{1}{2} )^2

Solving:

(-\frac{1}{2} )^3.(-\frac{1}{2} )^2\\=(-\frac{1}{8} ).(\frac{1}{4} )\\=-\frac{1}{32}

So, (-\frac{1}{2} )^3.(-\frac{1}{2} )^2=-\frac{1}{32}

4) \frac{2}{2^{-4}}

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

\frac{2}{2^{-4}}\\=2\times 2^4\\=2(16)\\=32

So, \frac{2}{2^{-4}} = 32

3 0
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It’d be 2
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