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Bingel [31]
3 years ago
10

Use the graph of the derivative of f to locate the critical points x0 at which f has neither a local maximum nor a local minimum

?
answer choices:
1. x0 = a , b
2. x0 = c , a
3. x0 = a , b , c
4. x0 = a
5. none of a , b , c
6. x0 = b
7. x0 = b , c
8. x0 = c

Mathematics
1 answer:
jok3333 [9.3K]3 years ago
6 0
<span>Critical points are where the derivative is 0, i.e. where it crosses the x - axis

The Critical points lies where the derivative is 0, while it crosses the x-axis, SO, in this case the choice 3 looks like best answer for this.
</span>
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Jim is an electrician. He charges a $150 home-visit fee and $45 per hour, with the first two hours free. Which equation could El
Vesna [10]
I might be wrong that the answer is D
5 0
3 years ago
-(1+7x)=6(-7-x)+36 <br> Alg 2
olganol [36]

Let's solve your equation step-by-step.

−(1+7x)=6(−7−x)+36

Step 1: Simplify both sides of the equation.

−(1+7x)=6(−7−x)+36

−7x+−1=(6)(−7)+(6)(−x)+36(Distribute)

−7x+−1=−42+−6x+36

−7x−1=(−6x)+(−42+36)(Combine Like Terms)

−7x−1=−6x+−6

−7x−1=−6x−6

Step 2: Add 6x to both sides.

−7x−1+6x=−6x−6+6x

−x−1=−6

Step 3: Add 1 to both sides.

−x−1+1=−6+1

−x=−5

Step 4: Divide both sides by -1.

−x

−1

=

−5

−1

x=5

Answer:

x=5

7 0
2 years ago
Toby wants to buy 2 concert tickets that cost $19 each. His makes $4.75 an hour while he mows lawns. How many hours will he need
nikitadnepr [17]
Cost of a ticket = $19
So, cost of two tickets = 19 * 2 = $38

Now, he makes, $4.75 in an hour.
Let, the number of hours for that much money = x

It would be: 4.75x = 38
x = 38 / 4.75
x = 8 hours

In short, Your Answer would be 8 hours

Hope this helps!
8 0
3 years ago
What is the 10th term in the pattern with the formula 5n + 100?
yanalaym [24]
Easy
first term is n=1
so 10th term is n=10

5(10)+100=50+100=150

150 is 10th term
5 0
3 years ago
A tank contains 1600 L of pure water. Solution that contains 0.04 kg of sugar per liter enters the tank at the rate 2 L/min, and
goldfiish [28.3K]

Let S(t) denote the amount of sugar in the tank at time t. Sugar flows in at a rate of

(0.04 kg/L) * (2 L/min) = 0.08 kg/min = 8/100 kg/min

and flows out at a rate of

(S(t)/1600 kg/L) * (2 L/min) = S(t)/800 kg/min

Then the net flow rate is governed by the differential equation

\dfrac{\mathrm dS(t)}{\mathrm dt}=\dfrac8{100}-\dfrac{S(t)}{800}

Solve for S(t):

\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{S(t)}{800}=\dfrac8{100}

e^{t/800}\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{e^{t/800}}{800}S(t)=\dfrac8{100}e^{t/800}

The left side is the derivative of a product:

\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/800}S(t)\right]=\dfrac8{100}e^{t/800}

Integrate both sides:

e^{t/800}S(t)=\displaystyle\frac8{100}\int e^{t/800}\,\mathrm dt

e^{t/800}S(t)=64e^{t/800}+C

S(t)=64+Ce^{-t/800}

There's no sugar in the water at the start, so (a) S(0) = 0, which gives

0=64+C\impleis C=-64

and so (b) the amount of sugar in the tank at time t is

S(t)=64\left(1-e^{-t/800}\right)

As t\to\infty, the exponential term vanishes and (c) the tank will eventually contain 64 kg of sugar.

7 0
3 years ago
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