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3 years ago

14

Noah and Andre are 15 miles apart on a bike path when they start biking toward each other . Noah rides at a constant speed of 4

miles per hour ,and Andre rides at a constant speed of 2 miles per hour . How long does it take until Noah and Andre meet?

1 answer:

In-s [12.5K]3 years ago

8 0

Answer:

jbejrnejnerbejdnedn

Step-by-step explanation:

njr3nirn3jrnmrn3kjrn3rj

rnheufenuenjebef\

kejirnrejfe

fjeifne

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Which expressions listed on the left are equivalent to √36a^8/225a^2? Check all that apply. (Assume that a ≠ 0.)

MAVERICK [17]

Answer: ABE

Step-by-step explanation:

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3 years ago

I will mark brainliesttt!

irinina [24]

Answer: 75 degrees

Step-by-step explanation: supplementary means 180 so you do 180-105 which is 75

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3 years ago

Find the inverse of f(x)=1/(x^3)

Pepsi [2]

Answer:

Step-by-step explanation:

y $f(x)^{-1} = inverse\\f(x)=y \\y = 1/(x^{3} \\Inverse: y=x ------------> x = 1/y^{3}\\y^{3} - \frac{1}{x} = 0\\y^{3} = \frac{1}{x}\\y = \sqrt[3]{\frac{1}{x}} \\y = \frac{\sqrt[3]{1} }{\sqrt[3]{x}} \\y = \frac{1}{\sqrt[3]{x}}$

4 0

3 years ago

Consider the following region R and the vector field F. a. Compute the two-dimensional curl of the vector field. b. Evaluate bo

Shalnov [3]

Looks like we're given

$\vec F(x,y)=\langle-x,-y\rangle$

which in three dimensions could be expressed as

$\vec F(x,y)=\langle-x,-y,0\rangle$

and this has curl

$\mathrm{curl}\vec F=\langle0_y-(-y)_z,-(0_x-(-x)_z),(-y)_x-(-x)_y\rangle=\langle0,0,0\rangle$

which confirms the two-dimensional curl is 0.

It also looks like the region $R$ is the disk $x^2+y^2\le5$ . Green's theorem says the integral of $\vec F$ along the boundary of $R$ is equal to the integral of the two-dimensional curl of $\vec F$ over the interior of $R$ :

$\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\iint_R\mathrm{curl}\vec F\,\mathrm dA$

which we know to be 0, since the curl itself is 0. To verify this, we can parameterize the boundary of $R$ by

$\vec r(t)=\langle\sqrt5\cos t,\sqrt5\sin t\rangle\implies\vec r'(t)=\langle-\sqrt5\sin t,\sqrt5\cos t\rangle$

$\implies\mathrm d\vec r=\vec r'(t)\,\mathrm dt=\sqrt5\langle-\sin t,\cos t\rangle\,\mathrm dt$

with $0\le t\le2\pi$ . Then

$\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\int_0^{2\pi}\langle-\sqrt5\cos t,-\sqrt5\sin t\rangle\cdot\langle-\sqrt5\sin t,\sqrt5\cos t\rangle\,\mathrm dt$

$=\displaystyle5\int_0^{2\pi}(\sin t\cos t-\sin t\cos t)\,\mathrm dt=0$

7 0

3 years ago

Please help me solve this equation and show the work show me how you got the answer

Nesterboy [21]

Answer: k=3

Step-by-step explanation:

To solve for k, you want to isolate k.

$\frac{2}{5} (5k+35)-8=12$ [add both sides by 8]

$\frac{2}{5} (5k+35)=20$ [multiply both sides by 5/2]

$5k+35=50$ [subtract both sides by 35]

$5k=15$ [divide both sides by 5]

$k=3$

Now, we know that k=3.

7 0

3 years ago