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3 years ago

Differentiate y=(6x)/(1-cot(x))

1 answer:

8 0

$\bf y=\cfrac{6x}{1-cot(x)}\implies \cfrac{dy}{dx}=\stackrel{quot ient~ru le}{\cfrac{6[1-cot(x)]~-~6x[-csc^2(x)]}{[1-cot(x)]^2}}
\\\\\\
\cfrac{dy}{dx}=\cfrac{6-6cot(x)~+~6xcsc^2(x)}{[1-cot(x)]^2}$

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Serena’s average speed during her training run was 8 kilometers per hour over the 2 hours that she ran. What was the total dista

Volgvan

Step-by-step explanation:

so, we just enter the given data into the given formula. and then calculate. it does not get easier than that.

d = 8km/h × 2h = 16km

Serena ran 16km in these 2 hours.

7 0

2 years ago

Please help asap!! Thank you

djyliett [7]

Step-by-step explanation:

Hey, there!!

Let's simply work with it,

60° + 2x° + (2x+12)° = 180° { being a linear pair}.

or, 72° + 4x = 180°

4x = 180° - 70°

$x = \frac{108}{4}$

Therefore, the value of x is 27°.

Now,

2x° = (2×27)°= 54°

(2x+12)° = 2×27°+12°= 66°.

(x+2)= 27+ 2= 29°

Now, The left angle is 27°.

Therefore, The answer is option A.

Hope it helps...

6 0

3 years ago

Daily low temperatures in Columbus, OH in January 2014 were approximately normally distributed with a mean of 15.45 and a standa

Alona [7]

Answer: 12.10

Step-by-step explanation:

Given : Mean : $\mu = 15.45$

Standard deviation : $\sigma = 13.70$

The formula to calculate the z-score :-

$z=\dfrac{x-\mu}{\sigma}$

For x= 5 degrees

$z=\dfrac{5-15.45}{13.70}=-0.7627737226\approx-0.76$

For x= 10 degrees

$z=\dfrac{10-15.45}{13.70}=-0.397810218\approx-0.40$

The P-value : $P(-0.76$

$=0.3445783-0.2236273=0.120951\approx0.1210$

In percent , $0.1210\times100=12.10\%$

Hence, the percentage of days had a low temperature between 5 degrees and 10 degrees = 12.10%

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2 years ago

Need help on num 15

Yakvenalex [24]

Answer:

Step-by-step explanation:

D's equation shows y equals 7.8 times x. 7.8 is the unit rate.

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3 years ago

Need answers as soon as possible please.

AysviL [449]

B
explanation: cuz i said so

you're welcome

3 0

1 year ago

Read 2 more answers