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3 years ago

How does the graph of y = 3x compare to the graph of y = 3–x?

Mathematics

2 answers:

bekas [8.4K]3 years ago

7 0

<h2>Answer and explanation:</h2>

In this problem, we have two functions. Let's call the first functions as:

$y=f(x)=3x$

and the second function:

$y=g(x)=3-x$

So we can compare this functions as follows:

f(x) has a positive slope while g(x) has a negative slope.

f(x) passes through the origin of the system of coordinates while g(x) cuts the y-intercept at $y=3$

g(x) has an x-intercept at $x=3$ while f(x) cuts the origin.

olga nikolaevna [1]3 years ago

6 0

Answer:

Both graphs represent the equation of a line

Step-by-step explanation:

Both graphs represent the equation of a straight line with the equation:

y = mx + c

Where m = the slope of the line and

c = intercept

Comparing both equations to the equation of a line given above:

In the first equation y = 3x

The slope, m = 3 (Positive slope)and

Intercept, c = 0

In the equation y = 3 -x

the slope, m = -1 (negative slope)

Intercept, c = 3

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If <img src="https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20x%20%3D%20log_%7Ba%7D%28bc%29" id="TexFormula1" title="\rm \: x = log_{a}(

timama [110]

Use the change-of-basis identity,

$\log_x(y) = \dfrac{\ln(y)}{\ln(x)}$

to write

$xyz = \log_a(bc) \log_b(ac) \log_c(ab) = \dfrac{\ln(bc) \ln(ac) \ln(ab)}{\ln(a) \ln(b) \ln(c)}$

Use the product-to-sum identity,

$\log_x(yz) = \log_x(y) + \log_x(z)$

to write

$xyz = \dfrac{(\ln(b) + \ln(c)) (\ln(a) + \ln(c)) (\ln(a) + \ln(b))}{\ln(a) \ln(b) \ln(c)}$

Redistribute the factors on the left side as

$xyz = \dfrac{\ln(b) + \ln(c)}{\ln(b)} \times \dfrac{\ln(a) + \ln(c)}{\ln(c)} \times \dfrac{\ln(a) + \ln(b)}{\ln(a)}$

and simplify to

$xyz = \left(1 + \dfrac{\ln(c)}{\ln(b)}\right) \left(1 + \dfrac{\ln(a)}{\ln(c)}\right) \left(1 + \dfrac{\ln(b)}{\ln(a)}\right)$

Now expand the right side:

$xyz = 1 + \dfrac{\ln(c)}{\ln(b)} + \dfrac{\ln(a)}{\ln(c)} + \dfrac{\ln(b)}{\ln(a)} \\\\ ~~~~~~~~~~~~+ \dfrac{\ln(c)\ln(a)}{\ln(b)\ln(c)} + \dfrac{\ln(c)\ln(b)}{\ln(b)\ln(a)} + \dfrac{\ln(a)\ln(b)}{\ln(c)\ln(a)} \\\\ ~~~~~~~~~~~~ + \dfrac{\ln(c)\ln(a)\ln(b)}{\ln(b)\ln(c)\ln(a)}$

Simplify and rewrite using the logarithm properties mentioned earlier.

$xyz = 1 + \dfrac{\ln(c)}{\ln(b)} + \dfrac{\ln(a)}{\ln(c)} + \dfrac{\ln(b)}{\ln(a)} + \dfrac{\ln(a)}{\ln(b)} + \dfrac{\ln(c)}{\ln(a)} + \dfrac{\ln(b)}{\ln(c)} + 1$

$xyz = 2 + \dfrac{\ln(c)+\ln(a)}{\ln(b)} + \dfrac{\ln(a)+\ln(b)}{\ln(c)} + \dfrac{\ln(b)+\ln(c)}{\ln(a)}$

$xyz = 2 + \dfrac{\ln(ac)}{\ln(b)} + \dfrac{\ln(ab)}{\ln(c)} + \dfrac{\ln(bc)}{\ln(a)}$

$xyz = 2 + \log_b(ac) + \log_c(ab) + \log_a(bc)$

$\implies \boxed{xyz = x + y + z + 2}$

(C)

6 0

1 year ago

The masses mi are located at the points pi. find the moments mx and my and the center of mass of the system. m1 = 2, m2 = 5, m3

creativ13 [48]

Most graphing calculators will do weighted averages pretty easily. It is mostly a matter of data entry.

mx = -2
my = 10
(x, y) = (mx, my)/10 = (-0.2, 1)

6 0

3 years ago

Combine like terms to create an equivalent expression. -1/2(-3y+10)

andrew-mc [135]

Answer:

3/2y -5

Step-by-step explanation:

-1/2 (-3y+10)

Distribute

-1/2* -3y + -1/2*10

3/2y -5

5 0

3 years ago

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There are 5 red marbles and 12 silver marbles in a bag. You randomly choose one of the marbles. What is the probability of choos

AleksandrR [38]

So we know there are 5 red and 12 silver marbles, together making 16 marbles.

This would mean that there is a 5/17 chance of choosing a red marble.

When you do the math to convert 5/17 into a decimal, it turns into being a .29 chance.

Convert .29 into a decimal, and you know that you have a 29% chance of grabbing a red marble.

Hope this helped!

- Kathryn Irene

6 0

2 years ago

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Sonja [21]

Answer:

i hope it will help you

7 0

2 years ago

Read 2 more answers