If m < 0 and b > 0, the graph of y = mx + b does not pass through which quadrant?

Mathematics

2 answers:

stiks02 [169]3 years ago

8 0

Answer:

Quadrant III

Step-by-step explanation:

The attached picture shows graph of 4 such linear functions with the conditions given in the problem. ALL of them DO NOT pass through Quadrant III.

The graphs shown are of the functions:

$y=-2x+1$

$y=-3x+3$

$y=-x+0.5$

$y=-5x+2$

So, any linear function of the form $y=mx+b$ with $m$ and $b>0$ does not pass through Quadrant III. Answer choice 3 is correct.

wlad13 [49]3 years ago

8 0

The graph will not past at quadrant III. I think thats the best answer, hope that helps.

You might be interested in

A bag contains 6 blue counters and 4 red counters. I take two counters out at random. My

Step2247 [10]

Answer:

Possible outcome=6+4=10

P(red counter)=2/10=1/5

6 0

3 years ago

Describe a situation that models a linear pattern and then describe a situation that models a nonlinear pattern. Do not state wh

UNO [17]

While linear equations are always straight, nonlinear equations often feature curves.

3 0

3 years ago

Read 2 more answers

I'm not sure where to start and how to differentiate this.

Mazyrski [523]

The picture is not clear. let me assume y = (x^4)ln(x^3)

product rule :
d f(x)g(x) = f(x) dg(x) + g(x) df(x)

dy/dx = (x^4)d[ln(x^3)/dx] + d[(x^4)/dx] ln(x^3)
= (x^4)d[ln(x^3)/dx] + 4(x^3) ln(x^3)

look at d[ln(x^3)/dx]
d[ln(x^3)/dx]
= d[ln(x^3)/dx][d(x^3)/d(x^3)]
= d[ln(x^3)/d(x^3)][d(x^3)/dx]
= [1/(x^3)][3x^2] = 3/x
... chain rule (in detail)

end up with
dy/dx = (x^4)[3/x] + 4(x^3) ln(x^3)
= x^3[3 + 4ln(x^3)]

5 0

3 years ago

TU = 3, RS = 15, TW = 4, TR = 9; Assume that the sides of triangle TUW are proportional to the sides of triangle TRS. What is th

VMariaS [17]

Answer:

$WS=8\ units$