Advanced Placement (AP), 25.10.2021 22:20 anggar20

How was the Supreme Court different in the early 1800s (List and describe 5 differences)

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light-year

the distance that light travels in a vacuum in one year, equal to about 9.46 trillion km (5.88 trillion mi). light-years are used in measuring interstellar and intergalactic distances.

parsec

a unit of distance used in astronomy, equal to about 3.25 light years (3.08 × 1016 meters).

parallax

the apparent angular displacement of a celestial body due to its being observed from the surface instead of from the center of the earth (diurnal parallax or geocentric parallax) or due to its being observed from the earth instead of from the sun (annual parallax or heliocentric parallax)

absolute magnitude

the magnitude (brightness) of a celestial object as it would be seen at a standard distance of 10 parsecs.

apparent magnitude

the magnitude of a celestial object as it is actually measured from the earth.

redshift

an increase in the wavelength of radiation emitted by a celestial body as a consequence of the doppler effect.

the great thing about your imagination is that it doesn't have to make sense! if i were you i would be stuck in space, no communication with earth travelling around and around, you can only guess how long its been since your communication broke down. you land on an undiscovered planet, you do a scientific investigation examining the elements and conditions. you come across alien lifeforms.

best of luck!

Answer from: Quest

ek = (m·v2) / 2

v - velocity of system after the collision

m = m1 + m2

ek = (m1 + m2) ·v2/2

ek = (200 + 250) ·2,32/2 = 450·5,29/2 = 1190,25 j

Answer from: Quest

Answerresearch products and services before buying;

Answer from: Quest

The answer is (d) 7/2======we have $f(x) = \begin{cases} x & \text{ for } x \le 0 \\ x + 1 & \text{ for } x \ \textgreater \ 0 \end{cases}$ we also have an integral $\displaystyle\int_{-2}^1 x\, f(x)\; dx$ notice how function f splits up at x = 0. there's a left piece and a right pieceso we'll need to do the same thing with our integral: split it at x = 0we use a property of integrals that allows us to split up and sum: $\displaystyle\int_{-2}^1 x\, f(x)\; dx = \displaystyle\int_{-2}^0 x\, f(x)\; dx + \displaystyle\int_{0}^1 x\, f(x)\; dx$ notice that for the piecewise function f, the function has the equationf(x) = x when x ≤ 0. so for the integral $\int_{-2}^0 x\, f(x)\; dx$ , f(x) = x so it changes into $\begin{aligned} \int_{-2}^0 x\, f(x)\; dx & = \int_{-2}^0 x \cdot x\; dx \\ & = \int_{-2}^0 x^2\; dx \\ & = \left[\frac{x^3}{3}\right]_{-2}^0 & & \text{\footnotesize antidifferentiate} \\ & = 0 - \frac{(-2)^3}{3} \\ & = \frac{8}{3} \end{aligned}$ for the piecewise function f, the function has the equationf(x) = x+1 when x > 0. for the integral $\int_{0}^1 x\, f(x)\; dx$ , we have $\begin{aligned} \int_{0}^1 x\, f(x)\; dx & = \int_{0}^1 x(x+1)\; dx \\ & = \int_{0}^1 (x^2+x)\; dx \\ & = \left[ \frac{x^3}{3} + \frac{x^2}{2}\right]_0^1 \\ & = \frac{1}{3} + \frac{1}{2} - 0 = \frac{5}{6} \end{aligned}$ therefore, back to our splitted integral: $\begin{aligned} \displaystyle\int_{-2}^1 x\, f(x)\; dx & = \displaystyle\int_{-2}^0 x\, f(x)\; dx + \displaystyle\int_{0}^1 x\, f(x)\; dx \\ & = \frac{8}{3} + \frac{5}{6} \\ & = \frac{7}{2} \end{aligned}$
Can anyone me on this ap calc question? ?

Can anyone me on this ap calc question? ?

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How was the Supreme Court different in the early 1800s (List and describe 5 differences)...