Just want to be sure that my answers is right, if anyone can help. thanks

I GIVE THANKS AND 5 STARS<br> look at the screenshot, help?? Please tell me how you got it!!

IrinaVladis [17]

We have to find out the coordinates of two points to find out the slope.
according to the figure,
first point is x1 = 1 & y1 = 2
second point is x2 = 4 & y2 = 1

slope = (y2 - y1) / (x2 - x1)
= (1-2) / (4-1)
= -1 / 3

3 0

3 years ago

How many solutions does 6x−1=34−x have

aleksley [76]

Step-by-step explanation:

6x+x=34+1

7x=35

x=35/7

x=5

Hence this equation has only one solution

4 0

3 years ago

If Layla gets paid $948 a slow week . The employees Marco gets paid $9 an hour ,Stephanie gets paid $10.50 an hour ,and Destiny

deff fn [24]

Answer:

they all got paid 474$ all together

Step-by-step explanation:

destiny worked 12 hours and made 240.17$

Stephanie worked 12 hours and made 126.17$

Marco worked 12 hours and made 108.17$

this all adds up to 474.50$ which is exactly half of 949$

5 0

3 years ago

Please help with the attached question. Thanks

Mademuasel [1]

Answer:

Choice A) $F(x) = 3\sqrt{x + 1}$ .

Step-by-step explanation:

What are the changes that would bring $G(x)$ to $F(x)$ ?

Translate $G(x)$ to the left by $1$ unit, and
Stretch $G(x)$ vertically (by a factor greater than $1$ .)

$G(x) = \sqrt{x}$ . The choices of $F(x)$ listed here are related to $G(x)$ :

Choice A) $F(x) = 3\;G(x+1)$ ;
Choice B) $F(x) = 3\;G(x-1)$ ;
Choice C) $F(x) = -3\;G(x+1)$ ;
Choice D) $F(x) = -3\;G(x-1)$ .

The expression in the braces (for example $x$ as in $G(x)$ ) is the independent variable.

To shift a function on a cartesian plane to the left by $a$ units, add $a$ to its independent variable. Think about how $(x-a)$ , which is to the left of $x$ , will yield the same function value.

Conversely, to shift a function on a cartesian plane to the right by $a$ units, subtract $a$ from its independent variable.

For example, $G(x+1)$ is $1$ unit to the left of $G(x)$ . Conversely, $G(x-1)$ is $1$ unit to the right of $G(x)$ . The new function is to the left of $G(x)$ . Meaning that $F(x)$ should should add $1$ to (rather than subtract $1$ from) the independent variable of $G(x)$ . That rules out choice B) and D).

Multiplying a function by a number that is greater than one will stretch its graph vertically.
Multiplying a function by a number that is between zero and one will compress its graph vertically.
Multiplying a function by a number that is between $-1$ and zero will flip its graph about the $x$ -axis. Doing so will also compress the graph vertically.
Multiplying a function by a number that is less than $-1$ will flip its graph about the $x$ -axis. Doing so will also stretch the graph vertically.

The graph of $G(x)$ is stretched vertically. However, similarly to the graph of this graph $G(x)$ , the graph of $F(x)$ increases as $x$ increases. In other words, the graph of $G(x)$ isn't flipped about the $x$ -axis. $G(x)$ should have been multiplied by a number that is greater than one. That rules out choice C) and D).

Overall, only choice A) meets the requirements.

Since the plot in the question also came with a couple of gridlines, see if the points $(x, y)$ 's that are on the graph of $F(x)$ fit into the expression $y = F(x) = 3\sqrt{x - 1}$ .