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kondaur [170]
3 years ago
7

What is the answer to a line GH passes through points (2,5) and (6,9). Wich equation represents line GH

Mathematics
1 answer:
Vadim26 [7]3 years ago
5 0
The answer is y = x + 3
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Is writing a 3 before a square root symbol the same as writing it after a square root symbol?
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This depends as to whether you are just multiplying the square root by 3, or are doing a cube root function. In a cube root, the 3 would be located just at the bent part of the square root symbol, usually in a smaller font. If one just wanted to multiply a square root by 3, it does not matter whether or not the 3 comes before or after the root. 

A cube root is equal to a value to the power of (1/3).

I hope this helps
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Show that (n+3)2 - (n-3)2 is an even number for all positive interger values of n
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(n+3)^2-(n-3)^2=\\
n^2+6n+9-(n^2-6n+9)=\\
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3 years ago
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Which expression is equivalent to 7(a+2) when a= 6?<br> O 7(6)<br> 916)<br> 0 7(6)+2<br> 7(6)+ 14
bearhunter [10]

Answer:

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7(a + 2) = 7(6)+ 14

Step-by-step explanation:

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3 years ago
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The logarithm of a number to the base V2 is k. What is its logarithm to the base 2v2 ?​
kakasveta [241]

Answer:

log_{2\sqrt 2} X = \frac{1}{3}k

Step-by-step explanation:

Given

Let the number be X

From the first statement, we have:

log_{\sqrt 2} X = k

Required

Find log_{2\sqrt 2} X

log_{\sqrt 2} X = k

using the following law of logarithm

log_ab = n, b=a^n

So:

log_{\sqrt 2} X = k

X = \sqrt{2}^k

Substitute: X = \sqrt{2}^k in log_{2\sqrt 2} X

log_{2\sqrt 2} X = log_{2\sqrt 2} ( \sqrt{2}^k)

log_{2\sqrt 2} X = klog_{2\sqrt 2} \sqrt{2}

Apply the following law:

log_ab = \frac{log\ b}{log\ a}

log_{2\sqrt 2} X = k\frac{log\ \sqrt 2}{log\ {2\sqrt 2}}

Express the square roots as power

log_{2\sqrt 2} X = k\frac{log\ 2^\frac{1}{2}}{log\ {2 * 2^\frac{1}{2}}}

log_{2\sqrt 2} X = k\frac{log\ 2^\frac{1}{2}}{log\ {2^\frac{3}{2}}}

using the following law of logarithm

log_ab = n, b=a^n

log_{2\sqrt 2} X = k\frac{\frac{1}{2}log\ 2}{\frac{3}{2}log\ 2}}

log_{2\sqrt 2} X = k\frac{\frac{1}{2}}{\frac{3}{2}}}

Rewrite as:

log_{2\sqrt 2} X = k * \frac{1}{2} \div\frac{3}{2}

log_{2\sqrt 2} X = k * \frac{1}{2} *\frac{2}{3}

log_{2\sqrt 2} X = k * \frac{1}{1} *\frac{1}{3}

log_{2\sqrt 2} X = \frac{1}{3}k

4 0
3 years ago
Ƒ (x) = x² − 2x – 4.
Kisachek [45]

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2 years ago
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