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SashulF [63]
1 year ago
13

Find a first-degree polynomial function p1 whose value and slope agree with the value and slope of f at x = c. f(x) = cot(x), c

= 4
Physics
1 answer:
Neporo4naja [7]1 year ago
7 0

The correct answer is y=-2x+(1/2)

y = f'(x)· x + c

Y = -2x + C

1 = -2x π/4 + C

=) C = I + π/2

y=-2x+(1/2)  is the first-degree polynomial.

First-degree polynomials are the simplest polynomials. Here, we'll talk about a few qualities and connect the terms polynomial, function, and equation. Write a polynomial equation in standard form before attempting to solve it. Factor it, then set each variable factor to zero after it has reached zero. The original equations' answers are the solutions to the derived equations. Factoring cannot always be used to solve polynomial equations. For instance, the polynomial 2x+5 has an exponent of 1. The most typical kinds of polynomials used in algebra and precalculus are zero polynomial functions.

Learn more about polynomial functions here :-

brainly.com/question/22592200

#SPJ4

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In 2007, michael carter (u.s.) set a world record in the shot put with a throw of 24.77 m. what was the initial speed of the sho
Serga [27]

The initial speed of the shot is 15.02 m/s.

The Shot put is released at a height y<em> </em>from the ground with a speed u. It is released at an angle θ to the horizontal. In a time t, the shot put travels a distance <em>R</em> horizontally.

Pl refer to the attached diagram.

Resolve the velocity u into horizontal and vertical components, u ₓ=ucosθ and uy=u sinθ. The horizontal component remains constant in the absence of air resistance, while the vertical component varies due to the action of the gravitational force.

Write an expression for R.

R=u_xt=(ucos \theta)t

Therefore,

t=\frac{R}{ucos\theta} .......(1)

In the time t, the net displacement of the shotput is y in the downward direction.

Use the equation of motion,

y=u_yt-\frac{1}{2}gt^2=(usin\theta) t-\frac{1}{2}gt^2

Substitute the value of t from equation (1).

y=(ucos\theta)(\frac{R}{ucos\theta} )-\frac{1}{2} g(\frac{R}{ucos\theta} )^2\\ =Rtan\theta-(\frac{gR^2}{2u^2cos^2\theta} )

Substitute -2.10 m for y, 24.77 m for R and 38.0° for θ and solve for u.

y=Rtan\theta-(\frac{gR^2}{2u^2cos^2\theta} )\\ (-2.10m)=(24.77 m)(tan38.0^o)-\frac{(9.8 m/s^2)(24.77m)^2}{2u^2(cos38.0^o)^2} \\ u^2=225.71(m/s)^2\\ u=15.02m/s

The shot put was thrown with a speed 15.02 m/s.




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