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2 years ago

Please help me out! I really need to get this done.

Mathematics

1 answer:

AlexFokin [52]2 years ago

3 0

A. Range (0,4)
B. Range (0,0)
A. Domain All real numbers
B. Domain all real numbers

i’m not too sure about the domain but I’m pretty sure it’s all real numbers because of the arrows at each end of the line

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I know you probably don’t see this but i forgot to do my homework it’s due tomorrow and it’s the called “Volume Unit Study Guide

saw5 [17]

Answer: Make sure all words are spelled correctly.

Try different keywords.

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Step-by-step explanation:

6 0

2 years ago

Need help Identify three solutions of the graph shown

tester [92]

The solutions are basically the points on the graph that the line passes through. The best way to pick them is to use the whole numbers that are on the corners of the little boxes instead of the middle.

(4, -5)
(3, -3)
(1, 1)
(0, 3)
(-1, 5)
and so on.

6 0

3 years ago

What is the inverse of the function? f(x)=3^x−1

geniusboy [140]

$y = f(x) = 3^{x} - 1$

let x be y and y be the x, so we have

$x = 3^{y} - 1$

$3^{y} = x + 1$

$log_{3}(3)^{y} = log_3(x + 1)$

$y = log_{3}(x+1)$

7 0

2 years ago

Convert each part of the ratio to percentage (a) 3:1 (b) 2:3:5 (c) 1:4 (d) 1:2:5

Scorpion4ik [409]

3:1 =300%
2:3:5 =50%
1:4 =25%
1:2:5 =62.5%

5 0

2 years ago

What is the equation of this graphed line?

barxatty [35]

Answer:

$y=-\frac{1}{3}x-5$

Step-by-step explanation:

Given:

The two points on the line are:

$(x_1,y_1)=(-6,-3)\\(x_2,y_2)=(6,-7)$

Now, for a line with two points on it, the slope of the line is given as:

$m=\frac{y_2-y_1}{x_2-x_1}$

For the points $(x_1,y_1)=(-6,-3)\ and\ (x_2,y_2)=(6,-7)$ , slope is:

$m=\frac{-7-(-3)}{6-(-6)}\\m=\frac{-7+3}{6+6}\\m=\frac{-4}{12}=-\frac{1}{3}$

Now, for a line with slope 'm' and a point $(x_1,y_1)$ on it is given as:

$y-y_1=m(x-x_1)$

Plug in all the values and determine the equation of the line. This gives,

$y-(-3)=-\frac{1}{3}(x-(-6))\\\\y+3=-\frac{1}{3}(x+6)\\\\y+3=-\frac{1}{3}x-\frac{1}{3}\times 6\\\\y+3=-\frac{1}{3}x-2\\\\y=-\frac{1}{3}x-2-3\\\\y=-\frac{1}{3}x-5$

Therefore, the equation of the line is:

$y=-\frac{1}{3}x-5$

4 0

3 years ago