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3 years ago

The length of arcXY is 48cm. What is the circumference of circle Z?

Mathematics

1 answer:

svp [43]3 years ago

8 0

Answer:

Step-by-step explanation:

In any circle the following ratios are always equal.

$\frac{arc}{circumference}$ = $\frac{centralangle}{360}$ , thus

$\frac{48}{circum}$ = $\frac{60}{360}$ = $\frac{1}{6}$ ( cross- multiply )

circumference = 6 × 48 = 288 cm → C

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Answer:

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Step-by-step explanation:

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Answer:

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3 years ago

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3 years ago

Find the differential of each function. (a) y = tan( 5t ) dy = Correct: Your answer is correct. (b) y = 5 − v2 5 + v2

Trava [24]

Answer:

$(a)$ $dy = 5sec^2(5t) \ dt$

$(a) \ dy = \frac{-20v}{(5+v^2)^2} \ dt$

Step-by-step explanation:

Given;

(a) y = tan(5t)

$(b) \ y = \frac{5-v^2}{5+v^2}$

Solving for (a)

y = tan(5t)

let u = 5t

⇒y = tan(u)

du/dt = 5

dy/du = sec²u

$\frac{dy}{dt} =\frac{dy}{du} *\frac{du}{dt} \\\\\frac{dy}{dt} = sec^2(u)*5\\\\\frac{dy}{dt} = 5sec^2(u)\\\\\frac{dy}{dt} = 5sec^2(5t)$

$dy = 5sec^2(5t) \ dt$

Solving for b;

let u = 5 - v²

du/dv = -2v

let v = 5+ v²

dv/du = 2v

$\frac{dy}{dv} = \frac{vdu - udv}{v^2} \\\\\frac{dy}{dv} = \frac{-2v(5+v^2) - 2v(5-v^2)}{(5+v^2)^2}\\\\\frac{dy}{dv} = \frac{-10v-2v^3-10v+2v^3}{(5+v^2)^2}\\\\$

$\frac{dy}{dv} = \frac{-10v-10v}{(5+v^2)^2}\\\\\frac{dy}{dv} = \frac{-20v}{(5+v^2)^2}\\\\dy = \frac{-20v}{(5+v^2)^2} dt$

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3 years ago

Expand the following expression −7(5x−8)

KonstantinChe [14]

-7(5x-8)
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