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3 years ago

Differentiate 2sin(5x-3)

Mathematics

1 answer:

anygoal [31]3 years ago

3 0

$f(x)=2\sin(5x-3)\\
f'(x)=2\cos(5x-3)\cdot5=10\cos(5x-3)$

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Given image A' B' C' D' E' if the pre-image contained point A(-1,5), which of the transformation resulted in image A' B' C' D' E

Svetradugi [14.3K]

Answer: c

Step-by-step explanation:

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4 years ago

An hourglass, composed of two identical cones, is 12 cm tall. The radius of each cone is 3 cm. If you want to fill the bottom ha

EleoNora [17]

The volume of a cone is V= (1/3)pi r²h
it was given
h=12cm
radius=3cm
the volume of the first cone is V1= 1/3)pi r²h1
where h1= h/2=6cm
so its volume is V1=1/3)xpix3²x6=56.52cm^3

and since the two cones are identical so V1=V2=56.52cm^3,
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4 years ago

In ΔOPQ, the measure of ∠Q=90°, the measure of ∠O=26°, and QO = 4.9 feet. Find the length of PQ to the nearest tenth of a foot.

Step2247 [10]

Given:

In ΔOPQ, m∠Q=90°, m∠O=26°, and QO = 4.9 feet.

To find:

The measure of side PQ.

Solution:

In ΔOPQ,

$m\angle O+m\angle P+m\angle Q=180^\circ$ [Angle sum property]

$26^\circ+m\angle P+90^\circ=180^\circ$

$m\angle P+116^\circ=180^\circ$

$m\angle P=180^\circ -116^\circ$

$m\angle P=64^\circ$

According to Law of Sines, we get

$\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$

Using the Law of Sines, we get

$\dfrac{p}{\sin P}=\dfrac{o}{\sin O}$

$\dfrac{QO}{\sin P}=\dfrac{PQ}{\sin O}$

Substituting the given values, we get

$\dfrac{4.9}{\sin (64^\circ)}=\dfrac{PQ}{\sin (26^\circ)}$

$\dfrac{4.9}{0.89879}=\dfrac{PQ}{0.43837}$

$\dfrac{4.9}{0.89879}\times 0.43837=PQ$

$2.38989=PQ$

Approximate the value to the nearest tenth of a foot.

$PQ\approx 2.4$

Therefore, the length of PQ is 2.4 ft.

4 0

3 years ago

X + 5y = 28 2x + 3y = 21

svp [43]

Answer: for first one x=28-5y

second one x=-3/29(y-7)

Step-by-step explanation:

im sure its right

4 0

3 years ago

A 10 foot ladder is leaning against a wall. Call x the distance from the top of the ladder to the ground, and call y the distanc

liq [111]

$-5.3\:\text{ft/s}$

Step-by-step explanation:

We start by applying the Pythagorean theorem to the ladder, with its length L as the hypotenuse:

$L^2 = 100\:\text{ft}^2 = x^2 + y^2$

where x is the vertical distance from the top of the ladder to the ground and y is the horizontal distance from the bottom of the ladder to the wall. Taking the derivative of the above expression with respect to time, we get

$0 = 2x\dfrac{dx}{dt} + 2y\dfrac{dy}{dt}$