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leonid [27]
3 years ago
8

a college student needs 11 classes that are worth a total of 40 credits in order to complete her degree. The college offers both

4-credit classes and 3-credit classes. Which system of equations can be used to determine f, the number of 4-credit classes the student can take to complete her degree, and h, the number of 3-credit classes? A) f + h = 40 4h + 3f = 11 B) f + h = 11 4h + 3f = 40 C) f + h = 40 4f + 3h = 11 D) f + h = 11 4f + 3h = 40
Mathematics
1 answer:
Rasek [7]3 years ago
8 0
I’m not entirely sure but I think it’s D
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Drupady [299]
It is between 2 and 3
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Which expression is equal to –1? cot StartFraction pi Over 4 EndFraction sin StartFraction pi over 2 EndFraction sin StartFracti
Bumek [7]

Given:

The four expression in the options.

To find:

Which expression is equal to –1.

Solution:

In option A,

\cot \left(\dfrac{\pi}{4}\right)=1

In option B,

\sin \left(\dfrac{\pi}{2}\right)=1

In option C,

\sin \left(\dfrac{3\pi}{2}\right)=\sin \left(\pi+\dfrac{\pi}{2}\right)

\sin \left(\dfrac{3\pi}{4}\right)=-\sin \left(\dfrac{\pi}{2}\right)   [\because \sin(\pi+\theta)=-\sin \theta]

\sin \left(\dfrac{3\pi}{4}\right)=-1

In option D,

\tan \left(\dfrac{5\pi}{4}\right)=\tan \left(\pi+\dfrac{\pi}{4}\right)

\tan \left(\dfrac{5\pi}{4}\right)=\tan \left(\dfrac{\pi}{4}\right)   [\because \tan(\pi+\theta)=\tan \theta]

\tan \left(\dfrac{5\pi}{4}\right)=1

Therefore, the correct option is C.

7 0
3 years ago
A solid lies between planes perpendicular to the x-axis at x tox=. The cross sections perpendicular to the x-axis are circular d
cluponka [151]

It's unclear what the planes are supposed to be, so I'll take x=a and x=b with 0\le a.

The cross sections are disks with diameter \csc x-\cot x, so each disk of thickness \Delta x has a volume of

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Then taking infinitesimally thin disks, we find the solid has a volume of

\displaystyle\frac\pi4\int_a^b(\csc x-\cot x)^2\,\mathrm dx

Since

(\csc x-\cot x)^2=2\csc^2x-2\csc x\cot x-1

and

\dfrac{\mathrm d(\csc x)}{\mathrm dx}=-\csc x\cot x

\dfrac{\mathrm d(\cot x)}{\mathrm dx}=-\csc^2x

it follows that the volume is

\displaystyle\frac\pi4\left(-2\cot x+2\csc x-x\right)\bigg|_a^b

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=\dfrac\pi4\left(2\tan\dfrac b2-2\tan\dfrac a2+a-b\right)

3 0
3 years ago
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