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3 years ago

How many quarts does it take to fill a kitchen sink

1 answer:

3 0

The average kitchen sink has a width of approximately 23.5 inches, a length of approximately 31.5 inches, and a depth of approximately 21 inches.

Multiplying these values together, we can see that the volume of the average kitchen sink is about 15545.25 cubic inches.

The conversion of cubic inches to liquid gallons is 1 cubic inch = approximately 0.004329 gallons.

Multiplying these numbers, we can see that 15545.25 cubic inches is the same as approximately 67.29538725 gallons. Now, we just have to multiply that number by 4, since there are 4 quarts in a gallon.

67.29538725 * 4 = 269.181549

It takes approximately 269.181549 quarts to fill the average kitchen sink.
Hope that helped =)

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Given: 1; -5; -13 ; -23 ; ...<br><br> Derive a formula for the nth term in the pattern.

mylen [45]

Answer:

f(n) = -n^2 -3n +5

Step-by-step explanation:

Suppose the formula is ...

f(n) = an^2 +bn +c

Then we have ...

f(1) = 1 = a(1^2) +b(1) +c

f(2) = -5 = a(2^2) +b(2) +c

f(3) = -13 = a(3^2) +b(3) +c

Here's a way to solve these equations.

Subtract the first equation from the second:

-6 = 3a +b . . . . . 4th equation

Subtract the second equation from the third:

-8 = 5a +b . . . . . 5th equation

Subtract the fourth equation from the fifth:

-2 = 2a

a = -1

Then substituting into the 4th equation to find b, we have ...

-6 = 3(-1) +b

-3 = b

and ...

1 = -1 +(-3) +c . . . . . substituting "a" and "b" into the first equation

5 = c

The formula is ...

f(n) = -n^2 -3n +5

3 0

4 years ago

Find the 3rd term in the expansion of (a+b)^8 in simplest form.

Art [367]

Answer:

The fifth term is the middle term of nine, with coefficient given by all the ways of choosing 4 items out of 8 , namely the ways of choosing 4 a 's out of 8 binomial factors.

Step-by-step explanation:

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3 years ago

Which choices are equations for the line shown below? Check all that apply.

nata0808 [166]

Answer:

c and b

Step-by-step explanation:

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3 years ago

Complete the factored form,<br> 46x + 9x-2-(15%-2))

VladimirAG [237]

Answer:55x+

−3

Step-by-step explanation:

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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

2 years ago