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2 years ago

9(x+3)=7x-2x+4 solve forx

Mathematics

2 answers:

san4es73 [151]2 years ago

5 0

Answer:

x=-5.75

Step-by-step explanation:

9(x+3)=7x-2x+4

9x+27=5x+4

4x+27=4

4x=-23

4x/4=-23/4

x=-5.75

Paladinen [302]2 years ago

3 0

9x+12=5x+4
9x-5x=12-4
4x=8
x=8/4
x=2

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2 years ago

Given y = log3(x + 4), what is the range?

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The range is all real numbers.

The domain is all reals numbers that are greater than -4.

Step-by-step explanation:

$y=\log_3(x+4)$ only exists when $x+4$ is positive.

You can take the log of a negative or 0 number.

So $x+4>0$ implies $x>-4$ . (I just subtract 4 on both sides.)

So the domain is x>-4. You should see this also when you graph the curve that the curve only exist to the right of -4.

Now the range. The range is where the curve exist for the y-values.

The equivalent exponent form of $y=\log_3(x+4)$ is $3^{y}=x+4$

We can solve this for x be subtract 4 on both sides:

$x=3^y-4$

Now here y can be anything; there are no restrictions on the exponent.

Also if you look at the graph of $y=\log_3(x+4)$ you should see every y getting hit by the curve (look down to up; use the y-axis as a guide).

Let's think about the inverse I found above a little more (I'm going to swap x and y).

$y=3^x-4$ .

If we look at the domain and range of this we can just swap it to get the domain and range of $y=\log_3(x+4)$ .

$y=3^x-4$ is an exponential function of 3^x that has been moved down 4 units.

The range since it has been moved down 4 units is $(-4,\infty)$ .

The domain of an exponential function is all real numbers. There are no restrictions on what you can plug in for x.

So swapping these to find the domain and range of $y=\log_3(x+4)$ :

Domain: $(-4,\infty)$

Range : $(-\infty,\infty)$

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