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2 years ago

Solve the system of equations y = 6x -7 y = 3x + 11

Mathematics

1 answer:

max2010maxim [7]2 years ago

7 0

If they both are equal to y, they must be equal to each other therefore,
6x-7 = 3x+11
3x = 18
x=6

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Please help me<br> solve (x+3) (x+7)

vodka [1.7K]

Step-by-step explanation:

Here, we'll need to multiply these two values together. I'll use the expansion formula, which goes as follows:

$(a+b)(c+d)$

Expand

$ac + ad + bc + bd$

Lets apply this to the following equation:

$(x+3) (x+7)$

Expand.

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2 years ago

Simplify the expression below and write it as a single logarithm:

OLEGan [10]

The simplification of 3log(x + 4) – 2log(x – 7) + 5log(x - 2) - log(x^2) is $\log \left(\frac{(x+4)^{3} \times(x-2)^{5}}{(x-7)^{2} \times x^{2}}\right)$

<u>Solution:</u>

Given, expression is $3 \log (x+4)-2 \log (x-7)+5 \log (x-2)-\log \left(x^{2}\right)$

We have to write in as single logarithm by simplifying it.

Now, take the given expression.

$\rightarrow 3 \log (x+4)-2 \log (x-7)+5 \log (x-2)-\log \left(x^{2}\right)$

Rearranging the terms we get,

$\left.\rightarrow 3 \log (x+4)+5 \log (x-2)-2 \log (x-7)+\log \left(x^{2}\right)\right)$

$\text { since a } \times \log b=\log \left(b^{a}\right)$

$\rightarrow \log (x+4)^{3}+\log (x-2)^{5}-\left(\log (x-7)^{2}+\log \left(x^{2}\right)\right)$

$\text { We know that } \log a \times \log b=\log a b$

$\rightarrow \log \left((x+4)^{3} \times(x-2)^{5}\right)-\left(\log \left((x-7)^{2} \times\left(x^{2}\right)\right)\right.$

$\text { We know that } \log a-\log b=\log \frac{a}{b}$

$\rightarrow \log \left(\frac{(x+4)^{3} \times(x-2)^{5}}{(x-7)^{2} \times x^{2}}\right)$

Hence, the simplified form $\rightarrow \log \left(\frac{(x+4)^{3} \times(x-2)^{5}}{(x-7)^{2} \times x^{2}}\right)$

4 0

2 years ago

Read 2 more answers

Given the two functions,

gogolik [260]

Graph: 2x
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6 0

3 years ago