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1 year ago

Find an equation in x and y for the line tangent to the curve ()=4,()=cos() at the point where =4.

Physics

1 answer:

m_a_m_a [10]1 year ago

7 0

An equation in x and y for the line tangent to the curve ()=4,()=cos() at the point where =4 is x(t)=2t+2,y(t)=t^4.

<h3>What is tangent?</h3>

In calculation, the digression line to a plane bend at a given point is the straight line that "simply contacts" the bend by then. Leibniz characterized it as the line through a couple of boundlessly close focuses on the bend. The chart of digression is intermittent, implying that it rehashes the same thing endlessly. In contrast to sine and cosine in any case, digression has asymptotes isolating every one of its periods. The space of the digression capability is all genuine numbers with the exception of at whatever point cos⁡(θ)=0, where the digression capability is vague. Assuming they stroll in an orderly fashion, they are fundamentally following a digression way for the shape that is made inside the fencing.

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12345 [234]

Answer:

charcoal is the correct answer

hope it helps , pls mark me as brainliest

6 0

3 years ago

We know that the Moon revolves around Earth during a period of 27.3 days. The average distance from the center of Earth to the c

PtichkaEL [24]

Answer:

Explanation:

This is a circular motion questions

Where the oscillation is 27.3days

Given radius (r)=3.84×10^8m

Circular motion formulas

V=wr

a=v^2/r

w=θ/t

Now, the moon makes one complete oscillation for 27.3days

Then, one complete oscillation is 2πrad

Therefore, θ=2πrad

Then 27.3 days to secs

1day=24hrs

1hrs=3600sec

Therefore, 1day=24×3600secs

Now, 27.3days= 27.3×24×3600=2358720secs

t=2358720secs

Now,

w=θ/t

w=2π/2358720 rad/secs

Now,

V=wr

V=2π/2358720 ×3.84×10^8

V=1022.9m/s

Then,

a=v^2/r

a=1022.9^2/×3.84×10^8

a=0.0027m/s^2

3 0

3 years ago

One of the main factors driving improvements in the cost and complexity of integrated circuits (ICs) is improvements in photolit

nika2105 [10]

Answer:

0.000003782 m

0.000001891 m

0.000001197125 m

Explanation:

$\lambda$ = Wavelength = 248 nm

D = Diameter of beam = 1 cm

f = Focal length = 0.625 cm

The angle is given by

$\theta=\dfrac{1.22\lambda}{D}$

The width is given by

$d=2\theta f\\\Rightarrow d=2\dfrac{1.22\lambda f}{D}\\\Rightarrow d=2\dfrac{1.22\times 248\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.000003782\ m$