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3 years ago

Help with khan academy

Mathematics

2 answers:

Phantasy [73]3 years ago

8 0

C can be seen as it is located on the point 1.
B is the value which is greater than zero.
A is the value which is less than zero.
B is -2/7.
A is -4/7.
C is 0.

VladimirAG [237]3 years ago

4 0

I’ve attached my work, hope it helps, just flip a and b

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Gavin’s unpaid credit card balance is $2103.23. His APR is 14.4%.

insens350 [35]

Answer:

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Leroy took a trip to Alaska. His airfare was $243 and he spent $72 each day while

Aleks04 [339]

Answer:

Total cost of the trip = 243 + 72x

Step-by-step explanation:

Cost of airfare = $243

Amount spent per day = $72

Which expression can he use to help him find the cost of the trip?

Let

x = number of days spent in Alaska

Total cost of the trip = Cost of airfare + (Amount spent per day * number of days spent in Alaska)

= 243 + (72 * x)

= 243 + 72x

Total cost of the trip = 243 + 72x

If he spent 4 days on the trip

Total cost of the trip = 243 + 72x

= 243 + 72(4)

= 243 + 288

= $531

4 0

2 years ago

What is 20 - {4 + [4 + (10/2)]}

vampirchik [111]

Answer:

Step-by-step explanation:

Plugged it in the calc!

Hope it helps :)

3 0

3 years ago

Show ur work plz!<br><br><br><br><br><br><br><br> ............

Kipish [7]

28-53= l -25l
Answer is 25

8 0

3 years ago

Calculate s f(x, y, z) ds for the given surface and function. g(r, θ) = (r cos θ, r sin θ, θ), 0 ≤ r ≤ 4, 0 ≤ θ ≤ 2π; f(x, y, z)

Triss [41]

$g(r,\theta)=(r\cos\theta,r\sin\theta,\theta)\implies\begin{cases}g_r=(\cos\theta,\sin\theta,0)\\g_\theta=(-r\sin\theta,r\cos\theta,1)\end{cases}$

The surface element is

$\mathrm dS=\|g_r\times g_\theta\|\,\mathrm dr\,\mathrm d\theta=\sqrt{1+r^2}\,\mathrm dr\,\mathrm d\theta$

and the integral is

$\displaystyle\iint_Sx^2+y^2\,\mathrm dS=\int_0^{2\pi}\int_0^4((r\cos\theta)^2+(r\sin\theta)^2)\sqrt{1+r^2}\,\mathrm dr\,\mathrm d\theta$

$=\displaystyle2\pi\int_0^4r^2\sqrt{1+r^2}\,\mathrm dr=\frac\pi4(132\sqrt{17}-\sinh^{-1}4)$

###

To compute the last integral, you can integrate by parts:

$u=r\implies\mathrm du=\mathrm dr$

$\mathrm dv=r\sqrt{1+r^2}\,\mathrm dr\implies v=\dfrac13(1+r^2)^{3/2}$

$\displaystyle\int_0^4r^2\sqrt{1+r^2}\,\mathrm dr=\frac r3(1+r^2)^{3/2}\bigg|_0^4-\frac13\int_0^4(1+r^2)^{3/2}\,\mathrm dr$

For this integral, consider a substitution of

$r=\sinh s\implies\mathrm dr=\cosh s\,\mathrm ds$

$\displaystyle\int_0^4(1+r^2)^{3/2}\,\mathrm dr=\int_0^{\sinh^{-1}4}(1+\sinh^2s)^{3/2}\cosh s\,\mathrm ds$

$\displaystyle=\int_0^{\sinh^{-1}4}\cosh^4s\,\mathrm ds$

$=\displaystyle\frac18\int_0^{\sinh^{-1}4}(3+4\cosh2s+\cosh4s)\,\mathrm ds$

and the result above follows.

4 0

3 years ago