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grigory [225]
2 years ago
8

Álgebra 1solve 4( -15x - 2) + 8

Mathematics
1 answer:
jonny [76]2 years ago
3 0

Answer:

-60x

Step-by-step explanation:

We simply open up the bracket and collect similar terms

So we have;

4(-15x-2) + 8 as:

4(-15x) -2(4) + 8

-60x -8 + 8

= -60x

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For what values of the variables are the following expressions defined? 1. 5y+2 2. 18/y 3. 1/x+7 4. 2b/10−b Example: X&gt;7
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1. All real numbers

2. All real numbers except y = 0

3. All real numbers except x = -7

4. All real numbers except b = 10

Step-by-step explanation:

For any function to be defined at a particular value, it should not be <em>approaching to a value </em>\infty<em> or it should not give us the </em>\frac{0}{0}<em> (zero by zero) form </em> when the input is given to the function.

The value of function will depend on the denominator.

Now, let us consider the given functions one by one:

1. 5y+2

Here denominator is 1. So, it can not attain a value \infty or \frac{0}{0}<em> (zero by zero) form </em>

So, for all real numbers, the function is defined.

2.\ \dfrac{18}{y}

At y = 0, the value

At\ y =0,  \dfrac{18}{y} \rightarrow \infty

So, the given function is <em>defined for all real numbers except y = 0</em>

<em></em>

<em></em>3.\ \dfrac{1}{x+7}<em></em>

Let us consider denominator:

x + 7 can be zero at a value x = -7

At\ x =-7,  \dfrac{1}{x+7} \rightarrow \infty

So, the given function is <em>defined for all real numbers except x = -7</em>

<em></em>

4.\ \dfrac{2b}{10-b}

Let us consider denominator:

10-b can be zero at a value b = 10

At\ b =10,  \dfrac{2b}{10-b} \rightarrow \infty

So, the given function is <em>defined for all real numbers except b = 10</em>

<em></em>

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