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Lorico [155]
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?
Mathematics
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:

5 0
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

6 0
3 years ago
Find the inverse of f(x)=1/(x^3)
Pepsi [2]

Answer:

Step-by-step explanation:

yf(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
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