3 years ago

How is this solved please help anyone

Mathematics

1 answer:

irina [24]3 years ago

5 0

$h = 2 \\ 6x + 18 = 2(3x + 9) \\ 6x + 18 = 6x + 18 \\ 18 = 18 \: always \: true$

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Bond [772]

It’s equal to 1

2x divide by 2x = 1
— —
2x divide by 2x 1

which equals to 1

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OverLord2011 [107]

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All you gotta do is Multiply 36 with 98.20 and boom, you got the answer

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Shkiper50 [21]

Answer:

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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)

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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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3 years ago

Jim USA

mart [117]

none of your questions make sense

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