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Thepotemich [5.8K]
3 years ago
7

Please please help if u don’t know please don’t answer not trying to be mean I know u are trying

Mathematics
1 answer:
Gemiola [76]3 years ago
3 0

x= 3

awutydyt7asd67awt67

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36. Reduce to the lowest term <br>i) 477/636 <br>ii) 361/418​
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<h3> Answer:</h3><h3>The simplest form of 477/636 is 3/4.</h3><h3>The simplest form of 361/418 is 19/22</h3>

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Identify the domain and range of the relation:<br> (-5,-3), (-2,1), (1,4), (2,5)
jonny [76]

Domain is the x-coordinate and range is the y - coordinate

\therefore \: domain =  \{ - 5, \:  - 2, \: 1, \: 2 \} \\  \\ \&\: range =  \{  - 3, \: 1, \: 4, \: 5\}

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Which operations are the most efficient to use to solve the problem and what is the solution? Lucas is making a banner that has
ivolga24 [154]

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3 0
3 years ago
Please help meeeee math​
Sergeeva-Olga [200]

Q2. By the chain rule,

\dfrac{dy}{dx} = \dfrac{dy}{dt} \cdot \dfrac{dt}{dx} = \dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}

We have

x=2t \implies \dfrac{dx}{dt}=2

y=t^4+1 \implies \dfrac{dy}{dt}=4t^3

The slope of the tangent line to the curve at t=1 is then

\dfrac{dy}{dx} \bigg|_{t=1} = \dfrac{4t^3}{2} \bigg|_{t=1} = 2t^3\bigg|_{t=1} = 2

so the slope of the normal line is -\frac12. When t=1, we have

x\bigg|_{t=1} = 2t\bigg|_{t=1} = 2

y\bigg|_{t=1} = (t^4+1)\bigg|_{t=1} = 2

so the curve passes through (2, 2). Using the point-slope formula for a line, the equation of the normal line is

y - 2 = -\dfrac12 (x - 2) \implies y = -\dfrac12 x + 3

Q3. Differentiating with the product, power, and chain rules, we have

y = x(x+1)^{1/2} \implies \dfrac{dy}{dx} = \dfrac{3x+2}{2\sqrt{x+1}} \implies \dfrac{dy}{dx}\bigg|_{x=3} = \dfrac{11}4

The derivative vanishes when

\dfrac{3x+2}{2\sqrt{x+1}} = 0 \implies 3x+2=0 \implies x = -\dfrac23

Q4. Differentiating  with the product and chain rules, we have

y = (2x+1)e^{-2x} \implies \dfrac{dy}{dx} = -4xe^{-2x}

The stationary points occur where the derivative is zero.

-4xe^{-2x} = 0 \implies x = 0

at which point we have

y = (2x+1)e^{-2x} \bigg|_{x=0} = 1

so the stationary point has coordinates (0, 1). By its "nature", I assume the question is asking what kind of local extremum this point. Compute the second derivative and evaluate it at x=0.

\dfrac{d^2y}{dx^2}\bigg|_{x=0} = (8x-4)e^{-2x}\bigg|_{x=0} = -4 < 0

The negative sign tells us this stationary point is a local maximum.

Q5. Differentiating the volume equation implicitly with respect to t, we have

V = \dfrac{4\pi}3 r^3 \implies \dfrac{dV}{dt} = 4\pi r^2 \dfrac{dr}{dt}

When r=5\,\rm cm, and given it changes at a rate \frac{dr}{dt}=-1.5\frac{\rm cm}{\rm s}, we have

\dfrac{dV}{dt} = 4\pi (5\,\mathrm{cm})^2 \left(-1.5\dfrac{\rm cm}{\rm s}\right) = -150\pi \dfrac{\rm cm^3}{\rm s}

Q6. Given that V=400\pi\,\rm cm^3 is fixed, we have

V = \pi r^2h \implies h = \dfrac{400\pi}{\pi r^2} = \dfrac{400}{r^2}

Substitute this into the area equation to make it dependent only on r.

A = \pi r^2 + 2\pi r \left(\dfrac{400}{r^2}\right) = \pi r^2 + \dfrac{800\pi}r

Find the critical points of A.

\dfrac{dA}{dr} = 2\pi r - \dfrac{800\pi}{r^2} = 0 \implies r = \dfrac{400}{r^2} \implies r^3 = 400 \implies r = 2\sqrt[3]{50}

Check the sign of the second derivative at this radius to confirm it's a local minimum (sign should be positive).

\dfrac{d^2A}{dr^2}\bigg|_{r=2\sqrt[3]{50}} = \left(2\pi + \dfrac{1600\pi}{r^3}\right)\bigg|_{r=2\sqrt[3]{50}} = 6\pi > 0

Hence the minimum surface area is

A\bigg_{r=2\sqrt[3]{50}\,\rm cm} = \left(\pi r^2 + \dfrac{800\pi}r\right)\bigg|_{r=2\sqrt[3]{50}\,\rm cm} = 60\pi\sqrt[3]{20}\,\rm cm^2

Q7. The volume of the box is

V = 8x^2

(note that the coefficient 8 is measured in cm) while its surface area is

A = 2x^2 + 12x

(there are two x-by-x faces and four 8-by-x faces; again, the coefficient 12 has units of cm).

When A = 210\,\rm cm^2, we have

210 = 2x^2 + 12x \implies x^2 + 6x - 105 = 0 \implies x = -3 \pm\sqrt{114}

This has to be a positive length, so we have x=\sqrt{114}-3\,\rm cm.

Given that \frac{dx}{dt}=0.05\frac{\rm cm}{\rm s}, differentiate the volume and surface area equations with respect to t.

\dfrac{dV}{dt} = (16\,\mathrm{cm})x \dfrac{dx}{dt} = (16\,\mathrm{cm})(\sqrt{114}-3\,\mathrm{cm})\left(0.05\dfrac{\rm cm}{\rm s}\right) = \dfrac{4(\sqrt{114}-3)}5 \dfrac{\rm cm^3}{\rm s}

\dfrac{dA}{dt} = 4x\dfrac{dx}{dt} + (12\,\mathrm{cm})\dfrac{dx}{dt} = \left(4(\sqrt{114}-3\,\mathrm {cm}) + 12\,\mathrm{cm}\right)\left(0.05\dfrac{\rm cm}{\rm s}\right) = \dfrac{\sqrt{114}}5 \dfrac{\rm cm^2}{\rm s}

5 0
1 year ago
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