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jolli1 [7]
3 years ago
13

Jane read 3/8 of a book for 6 days. How many books did she read?

Mathematics
1 answer:
ZanzabumX [31]3 years ago
3 0

Answer:

2 and a half books.

Step-by-step explanation:

3/8×6=?

I simplified it by dividing 8 and 6 with 2 to get:

3/4×3=9/4

in whole numbers it will be, 2 and a half books

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The region in the first quadrant bounded by the x-axis, the line x = ln(π), and the curve y = sin(e^x) is rotated about the x-ax
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First, it would be good to know that the area bounded by the curve and the x-axis is convergent to begin with.

\displaystyle\int_{-\infty}^{\ln\pi}\sin(e^x)\,\mathrm dx

Let u=e^x, so that \mathrm dx=\dfrac{\mathrm du}u, and the integral is equivalent to

\displaystyle\int_{u=0}^{u=\pi}\frac{\sin u}u\,\mathrm du

The integrand is continuous everywhere except u=0, but that's okay because we have \lim\limits_{u\to0^+}\frac{\sin u}u=1. This means the integral is convergent - great! (Moreover, there's a special function designed to handle this sort of integral, aptly named the "sine integral function".)

Now, to compute the volume. Via the disk method, we have a volume given by the integral

\displaystyle\pi\int_{-\infty}^{\ln\pi}\sin^2(e^x)\,\mathrm dx

By the same substitution as before, we can write this as

\displaystyle\pi\int_0^\pi\frac{\sin^2u}u\,\mathrm du

The half-angle identity for sine allows us to rewrite as

\displaystyle\pi\int_0^\pi\frac{1-\cos2u}{2u}\,\mathrm du

and replacing v=2u, \dfrac{\mathrm dv}2=\mathrm du, we have

\displaystyle\frac\pi2\int_0^{2\pi}\frac{1-\cos v}v\,\mathrm dv

Like the previous, this require a special function in order to express it in a closed form. You would find that its value is

\dfrac\pi2(\gamma-\mbox{Ci}(2\pi)+\ln(2\pi))

where \gamma is the Euler-Mascheroni constant and \mbox{Ci} denotes the cosine integral function.
5 0
4 years ago
4y + 7x - 3y + 2x please help me answer thx
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4y+7x-3y+2x
 
(7x+2x)+(4y+-3y)

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The answers are C and B

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Write 8/5 in a mixed number
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3 0
3 years ago
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If MN=NQ and MQ=QR=RP, calculate for x​
Akimi4 [234]

Answer:

  • C. 18°

Step-by-step explanation:

  • <em>Refer to attached diagram</em>
<h3>Given </h3>
  • MN = NQ
  • MQ = QR = RP
  • NMR = 3x
  • QMR = x
<h3>To find</h3>
  • Value of x
<h3>Solution</h3>
  • MN = NQ  ⇒ ∠MRN = ∠QMN = 3x+x = 4x
  • MQ = QR ⇒ ∠MRQ = ∠QMR = x

<u>RQP is exterior angle of ΔMQR ⇒ </u>

  • ∠RQP = x + x = 2x

<u>QR = RP ⇒ </u>

  • ∠RPQ = ∠RQP = 2x

<u>∠QRN is exterior angle of ΔQRP ⇒ </u>

  • ∠QRN = 2x + 2x = 4x
  • ∠MRN = ∠QRN - ∠QRM = 4x - x = 3x

<u>∠MRN = ∠NMR = 3x ⇒ </u>

  • NR = MQ

<u>NR = MQ ⇒ </u>

  • ∠NQR = ∠QRN = 4x

<u>We now have a straight angle MQP:</u>

  • ∠MQP = ∠MQN + ∠RQN + ∠RQP
  • 180° = 4x + 4x + 2x
  • 10x = 180°
  • x = 18°

Correct choice is C

7 0
3 years ago
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