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2 years ago

Find the derivative of y=sin^(2)(x)cos^(2)(x)

1 answer:

5 0

$sin^{2}x \ cos^{2} x= \frac{1}{2} (1-cos2x) \times \frac{1}{2} (1+cos2x) \\ = \frac{1}{4} (1- cos^{2} 2x) \\ = \frac{1}{4} ( sin^{2} 2x +cos^{2} 2x- cos^{2} 2x) \\ = \frac{1}{4} ( sin^{2} 2x) \\ = \frac{1}{4} ( \frac{1}{2} (1-cos2(2x))) \\ = \frac{1}{8} (1-cos4x) \\ \frac{d}{dx} (sin^{2}x \ cos^{2} x)= \frac{1}{8} \frac{d}{dx} (1-cos4x) \\ = \frac{1}{8} [-4(-sin4x) \\ \frac{1}{2} sin4x$

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what is the equation of the line described below written in slope-intercept form? the line passing through point (2,2) and perpe

djverab [1.8K]

Answer:

The equation of line passing through point (2 , 2) and perpendicular to line y = x is y = - x + 4 .

Step-by-step explanation:

Given as :

The line equation is y = x

Now, equation of line in slope-intercept form y = m x + c

where m is the slope of line and c is y-intercept

Comparing given line equation with standard line equation

The slope of line y = x is m = 1

Again

Other line is passing through point (2 ,2) and is perpendicular to line y = x

Let The slope of other line = M

∵ From perpendicular lines property

Product of slope of lines = - 1

i.e m × M = - 1

Or , M = $\dfrac{ - 1}{m}$

Or, M = $\dfrac{ - 1}{1}$

∴ M = - 1

<u>Now, Equation of other line in point-slope form</u>

The other line is passing through point (2 , 2) and slope M = - 1

So, y - $y_1$ = M × (x - $x_1$ )

Or, y - 2 = - 1 × ( x - 2 )

Or, y - 2 = - x + 2

Or, y = - x + 2 + 2

Or, y = - x + 4

Hence, The equation of line passing through point (2 , 2) and perpendicular to line y = x is y = - x + 4 . Answer

5 0

3 years ago

Jon recently drove to visit his parents who live 280 miles away. On his way there his average speed was 9 miles per hour faster

Luba_88 [7]

Answer:

11 mph and 20 mph

Step-by-step explanation:

Represent his average speed going by r1 and his average speed returning by r2. We know that r1 = r2 + 9.

Recall that distance = rate times time, so time = distance / rate.

Time spent going was (280 mi) / r1, or (280 mi) / (r2 + 9 mph).

Time spend returning was (280 mi) / r2.

The total time was 14 hrs, so (280 mi) / (r2 + 9 mph) + (280 mi) / r2 = 14 hrs

Note that there is only one variable here: r2. Find r2, and then from r2, find r1:

Dividing all 3 terms by 14 hrs yields:

20 20

---------- + ----------- = 1

r2 + 9 r2

The LCD here is r2(r2 + 9). Thus, we have:

20r2 (r2 + 9)(r2)

------------------- = 1 or ------------------

(r2 + 9)(r2) (r2 + 9)(r2)

Then 20(r2) = (r2)^2 + 9(r2). This is reducible by dividing all terms by r2:

20 = r2 + 9, or 11 = r2. Then r1 = 11 + 9, or 20.

The two rates were 11 mph and 20 mph.

8 0

3 years ago

Math grade 9 , Aor B or C or D fassst plzzz

grigory [225]

Answer:

where are your questions?

Step-by-step explanation:

but you let me choice, I choice B.

6 0

2 years ago

Kurt drew a rectangular maze with the length of 3/4 foot and a width of 5/12 foot.

pochemuha

I hope this helps you

Area =3/4.5/12

Area =5/4.4

Area =5/16

4 0

3 years ago

Carina says that to divide by 12, she will need to write 12 as the fraction and then divide. Is she correct? Explain your reason

Alika [10]

Answer:

I do not know

Step-by-step explanation:

I think so

6 0

2 years ago