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Eddi Din [679]
3 years ago
14

Logan can read at a rate of 2/3 page per minute. At that rate, how many pages can he read in 2 hours?

Mathematics
1 answer:
Anon25 [30]3 years ago
7 0

Answer:

240- 360

Step-by-step explanation:

You have to find out how many minutes in an hour which is 60. Then multiply it by 2 for 2 hours to get 120. Then you multiply 2x120 to get 240. Then you multiply 3x120 to get 360.

I'm not sure if this is correct but this is how I would go about it. :)

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Options for awnser is a,b,c,d
Aleks04 [339]

Answer:

Step-by-step explanation:

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The right answer is D

We should rotate the R point 3/4 a circle clockwise so it falls on D point.

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3 years ago
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Pls help I’ve been getting bots all day! I really need help and I will give brainliest!
zaharov [31]

Not a robot! I don't think.

Y in the beginning goes up to 3.

Y in the end goes down to -2 before shooting back up in an infinite sense.

Increasing: The beginning and the end the line on the graph. (Also the jump in the middle, the round part.)

Decreasing: The middle of the graph. (The jump, downward slope.)

Constant, Y at the near end going in a straight line from 9-12 at a -2.

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3 years ago
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Explain to a Friend Taylor wants to buy a bicycle
dlinn [17]

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First add the 48.75 and 18.30 together.

second multiply 5.50 by 9

she would only earn 49.50 so she won’t be able to buy the bike and helmet

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2 years ago
What is HCF and LCM ? # confused
GaryK [48]
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3 years ago
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Find the following: F(x, y, z) = e^(xy) sin z j + y tan^−1(x/z)k Exercise Find the curl and the divergence of the vector field.
natulia [17]

\vec F(x,y,z)=e^{xy}\sin z\,\vec\jmath+y\tan^{-1}\dfrac xz\,\vec k

Divergence is easier to compute:

\mathrm{div}\vec F=\dfrac{\partial(e^{xy}\sin z)}{\partial y}+\dfrac{\partial\left(y\tan^{-1}\frac xz\right)}{\partial z}

\mathrm{div}\vec F=xe^{xy}\sin z-\dfrac{xy}{x^2+z^2}

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\mathrm{curl}\vec F=\begin{vmatrix}\vec\imath&\vec\jmath&\vec k\\D_x&D_y&D_z\\0&e^{xy}\sin z&y\tan^{-1}\frac xz\end{vmatrix}

\mathrm{curl}\vec F=\left(D_y\left[y\tan^{-1}\dfrac xz\right]-D_z\left[e^{xy}\sin z\right]\right)\,\vec\imath-D_x\left[y\tan^{-1}\dfrac xz\right]\,\vec\jmath+D_x\left[e^{xy}\sin z}\right]\,\vec k

\mathrm{curl}\vec F=\left(\tan^{-1}\dfrac xz-e^{xy}\cos z\right)\,\vec\imath-\dfrac{yz}{x^2+z^2}\,\vec\jmath+ye^{xy}\sin z\,\vec k

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3 years ago
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