Ask question

Ask question

All categories

3 years ago

13

Which equations have a leading coefficient of 3 and a constant term of –2? Check all that apply.

2 answers:

Svetllana [295]3 years ago

8 0

Answers are: choice A, choice C, choice E

Explanation: The leading coefficient is the number that is to the left of the term with the largest exponent, which in this case is 2. The term 3x^2 is the leading term with the coefficient of 3. This is why the leading coefficient is 3 for choice A, choice C, and choice E. Choice B has a leading coefficient of -3 so we can rule that out. Choice D has a leading coefficient of 3, but the constant term is NOT -2. Instead the constant term is +2, so we can rule out choice D as well. The other choices that haven't been eliminated all have 3x^2 somewhere in them, as well as the constant term -2. The other x term isn't relevant to the restrictions placed in the instructions.

Phoenix [80]3 years ago

5 0

Answer:

a,c,e

Step-by-step explanation:

You might be interested in

Solve the proportion for the variable.<br><br> W/5=49/35

hammer [34]

Answer:

7

Step-by-step explanation:

So if w/5 = 49/35, then you can take w/5 and make the denominator 35 by multiplying the top and bottom by 7:

w/5 * 7/7 = 7w/35

So then you have:

7w/35 = 49/35

Which means:

7w = 49

w = 7

7 0

3 years ago

Answer the following questions about the function whose derivative is f primef′(x)equals=x Superscript negative three fifths Ba

Elden [556K]

Answer:

(a) The critical points of f are x=0 and x=3.

(b)f is decreasing on $(0,3)$ and f is decreasing on $(3,\infty)$ .

(c) Therefore the local minimum of f is at x=3

Step-by-step explanation:

Given function is

$f'(x)= x^{-\frac35}(x-3)$

(a)

To find the critical point set f'(x)=0

$\therefore x^{-\frac35}(x-3)=0$

$\Rightarrow x=0,3$

The critical points of f are 0,3.

(b)

The interval are $(0,3)$ and $(3,\infty)$ .

To find the increasing or decreasing, taking two points one point from the interval (0,3) and another point $(3,\infty)$ .

Assume 1 and 4.

Now $f'(1)=(1)^{-\frac35}(1-3)$

and $f'(4)=(4)^{-\frac35}(4-3)>0$

Since 1∈ $(0,3)$ , f'(x)<0 and 4∈ $(3,\infty)$ , f'(x)>0

∴f is decreasing on $(0,3)$ and f is decreasing on $(3,\infty)$ .

(c)

$f'(x)= x^{-\frac35}(x-3)$

Differentiating with respect to x

$f''(x)=-\frac35x^{-\frac 85}(x-3)+x^{-\frac35}$

Now

$f''(0)=-\frac35(0)^{-\frac 85}(0-3)+(0)^{-\frac35}=0$

and

$f''(3)=-\frac35(3)^{-\frac 85}(3-3)+3^{-\frac35}$

$=0.517>0$

Since f''(x)>0 at x=3

Therefore the local minimum of f is at x=3

8 0

3 years ago

PLAESE HELP ME FAST THIS IS WORTH 20 POINTS

mojhsa [17]

Where is the question

3 0

4 years ago

Read 2 more answers

Troy is searching for a new home. He’ll need to take out a mortgage once he finds a house. Before visiting the bank, Troy decide

max2010maxim [7]

Answer:

Troy could have been a victim of identity theft , and he should check his credit report more often to avoid being surprised in the future.

Step-by-step explanation:

5 0

3 years ago

Kody’s alarm rings every 11 minutes, and his dad calls upstairs every 15 minutes to wake him up. But Kody will only wake up if h

IrinaK [193]

Answer:

10:15

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers