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rusak2 [61]
3 years ago
13

Dylan works out and burns 10 calories per minute swimming and 7 calories per minute biking. He wants to burn a minimum of 600 ca

lories. Due to daylight limitations, the time Dylan spends swimming must be no less than twice the time he spends biking.

Mathematics
1 answer:
Alexxandr [17]3 years ago
5 0

Answer:

The solution in the attached figure

Step-by-step explanation:

Let

x -----> the time Dylan spends swimming

y -----> the time Dylan spends biking

we know that

10x+7y \geq 600 -----> inequality A

x \geq 2y -----> inequality B

Solve the system by graphing

The solution is the shaded area

see the attached figure

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Help plsss; what is the inverse of f(x) = 5x - 1
Blababa [14]

Answer:

f^-1(x) = 1/5x + 1/5

Step-by-step explanation:

Switch the x and y values:

y = 5x - 1

x = 5y - 1

Put the equation back into y = form:

x = 5y - 1

x + 1 = 5y

1/5x + 1/5 = y

y = 1/5x + 1/5

f^-1(x) = 1/5x + 1/5

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3 years ago
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Answer: t=−3.7

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3 years ago
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CAN SOMEONE HELP ME IN THIS INTEGRAL QUESTION PLS
finlep [7]

Due to the symmetry of the paraboloid about the <em>z</em>-axis, you can treat this is a surface of revolution. Consider the curve y=x^2, with 1\le x\le2, and revolve it about the <em>y</em>-axis. The area of the resulting surface is then

\displaystyle2\pi\int_1^2x\sqrt{1+(y')^2}\,\mathrm dx=2\pi\int_1^2x\sqrt{1+4x^2}\,\mathrm dx=\frac{(17^{3/2}-5^{3/2})\pi}6

But perhaps you'd like the surface integral treatment. Parameterize the surface by

\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+u^2\,\vec k

with 1\le u\le2 and 0\le v\le2\pi, where the third component follows from

z=x^2+y^2=(u\cos v)^2+(u\sin v)^2=u^2

Take the normal vector to the surface to be

\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial u}=-2u^2\cos v\,\vec\imath-2u^2\sin v\,\vec\jmath+u\,\vec k

The precise order of the partial derivatives doesn't matter, because we're ultimately interested in the magnitude of the cross product:

\left\|\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right\|=u\sqrt{1+4u^2}

Then the area of the surface is

\displaystyle\int_0^{2\pi}\int_1^2\left\|\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right\|\,\mathrm du\,\mathrm dv=\int_0^{2\pi}\int_1^2u\sqrt{1+4u^2}\,\mathrm du\,\mathrm dv

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7 0
3 years ago
Can someone please help. I don’t know how to do it.
ira [324]

Answer:

152 m^2

Step-by-step explanation:

Find the area of the triangle and subtract it from the area of the square

area of half-triangle = 1/2 x base x height = 1/2 x 8 x 13 = 52

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area of the square = 16 x 16 = 256

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4 0
4 years ago
Evaluate the expression shown below and write your answer as a fraction in simplest form.
Ierofanga [76]

Answer:

72 / 5

Step-by-step explanation:

( 12 / 5 ) + ( 6 / 5 )

= ( 12 + 6 ) / 5

= 18 / 5

[ ( 12 / 5 ) + ( 6 / 5 ) ] / ( 1 / 4 )

= ( 18 / 5 ) / ( 1 / 4 )

= ( 18 x 4 ) / 5

= 72 / 5

8 0
3 years ago
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