Please please help mark as branliest

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

lora16 [44]

Answer:

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 : $(-\infty,\infty)$

4 0

3 years ago

A company charges $13 plus $3 per hour to rent a boat. Abigail and Monique want to rent a boat but do not want to spend more tha

Marat540 [252]

Answer:

13 + 3x = 20

Step-by-step explanation

3 0

3 years ago

The measures of two supplementary angles have a ratio of 4:5. What is the measure of the larger angle?

vladimir1956 [14]

Answer:

100 is larger, that is your answer.

Step-by-step explanation:

First, we know that all supplementary angles add to 180°. Then, since the angles are in the ratio 4:5, we can also represent these angles as 4x and 5x respectively. Both angle measures have the same variable (x) because no matter how large the angles are, they must have a common multiple because they are in a ratio. We can then write the formula 4x+5x=180 and solve for x. First, you combine like terms to get 9x=180. The you divide both sides by 9 to get x=20. Then, we can find both angles to be 80 (4*20) and 100 (5*20) by plugging them into 4x and 5x. Since 100 is larger, that is your answer.

I hope this helps.

7 0

4 years ago

Help me plz !!!!! Plz plz

Serjik [45]

Total weight of the steel gate = 150 kg

Solution:

Length = 12 m

Breadth = 5 m

Using Pythagoras theorem,

Diagonal = $\sqrt{l^2+w^2}$

$=\sqrt{12^2+5^2}$

$=\sqrt{144+25}$

$=\sqrt{169}$

Diagonal = 13 m

Total length of the gate = 5 + 5 + 12 + 12 + 13 + 13

Total length of the gate = 60 m

Weight of the steel per meter = 2.5 kg

Weight of the steel for 60 m = 60 × 2.5

= 150 kg

Total weight of the steel gate = 150 kg

8 0

4 years ago

ANSWER AND EXPLAIN I’LL MARK BRAINLIEST PLEASE ILL MARK BRAINLIEST

masya89 [10]

Answer:

160 meters

Step-by-step explanation:

Before I begin, note that distance = rate * time (or d = r * t)

To solve this problem, we need to find what point these two people have the same distance. (so Janiyah covers the same distance as Connor)

So, let's set the time that Janiyah takes as x (in seconds). So, he covers

d = r * t, or 8x meters

Then, Connor's equation would be

d = r * t

d = 5x + 60 (because he got a 60 meter headstart)

Because d is the same in both equations, to solve for the amount of time they run:

8x = 5x + 60

Subtract 5x from both sides.

3x = 60

x = 20

Then, to find Janiyah's distance (you can also use Connors because the two distances are equal), put x = 20 into 8x

8x = 8 * 20 = 160

So, Janiyah ran 160 meters.

Just in case, let's check if this is correct. Connor's distance equals Janiyah's, so

5x + 60 = 160

5 * 20 + 60 = 160

100 + 60 = 160

160 = 160

True.

So, the answers are correct.

I hope this helps! Feel free to ask any questions!

3 0

3 years ago