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3 years ago

Using the following image, find the. midpoint of the line by completing the problems attached.

Mathematics

2 answers:

enot [183]3 years ago

5 0

Answer:4/1 (I think)

Step-by-step explanation:

It’s rise over run, so if you look at it what’s the rise what’s the run, of the middle point

Aleonysh [2.5K]3 years ago

5 0

Answer:-3,1

Step-by-step explanation:

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What is the cubed root of 125? Please include step-by-step

denis23 [38]

Answer:

Step-by-step explanation:

cubes means 3. You do 5 to the 3rd power and you get 125

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grigory [225]

1/8 i think it is good bye

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Paladinen [302]

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3 years ago

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Find the differential of each function. (a) y = tan( 7t ) dy = Incorrect: Your answer is incorrect. (b) y = 4 − v2 4 + v2 dy =

Kisachek [45]

Answer:

$(a) dy= 7sec^2(7t)dt\\\\(b) dy = \frac{-16v}{(4+v^2)^2} dv$

Step-by-step explanation:

Given;

(a) y = tan (7t)

let u = 7t

y = tan(u)

du/dt = 7

dy/du = sec²(u)

$\frac{dy}{dt} = \frac{dy}{du} *\frac{du}{dt}\\\\\frac{dy}{dt} = sec^2(u)* 7\\\\\frac{dy}{dt} =7sec^2(u)\\\\\frac{dy}{dt} = 7sec^2(7t)\\\\dy = 7sec^2(7t)dt$

(b)

$y = \frac{4-v^2}{4+v^2}\\\\$

let u = 4 - v²

du/dv = -2v

let v = 4 + v²

dv/du = 2v

$y = \frac{4-v^2}{4+v^2}\\\\\frac{dy}{dv} = \frac{vdu-udv}{v^2} \\\\\frac{dy}{dv} =\frac{-2v(4+v^2)-2v(4-v^2)}{(4+v^2)^2}\\\\\frac{dy}{dv} =\frac{-8v-2v^3-8v+2v^3}{(4+v^2)^2}\\\\\frac{dy}{dv} =\frac{-16v}{(4+v^2)^2}\\\\dy = \frac{-16v}{(4+v^2)^2}dv$

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4 years ago