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zhannawk [14.2K]
3 years ago
6

Miguel is reading a 338-page book. In the first 40 minutes he spends reading, he reads 52 pages. If he continues reading at the

same rate, how many minutes will it take him to read the entire book?
Mathematics
2 answers:
enyata [817]3 years ago
6 0

Answer:

260 minutes

Step-by-step explanation:

40:52

x:338

338*40=52x

13,520/52

<em><u>260</u></em>:338

viktelen [127]3 years ago
4 0

Answer: 260 minutes.

Step-by-step explanation:

This represents the ratio between the minutes he reads and the pages he reads: 40 : 52

The book Miguel is reading is 338 pages long. Miguel also reads 52 pages in 40 minutes. You can divide 338 by 52, which is equal to 6.5.

If it takes Miguel 40 minutes to read 52 pages, then we can multiply that by 6.5. 40 x 6.5 = 260.

In conclusion, it will take Miguel 260 minutes to read the entire book, or 4 hours and 20 minutes.

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521,400 meters squared

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Step-by-step explanation

6 0
3 years ago
Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y
irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

y'(x)= x^2y(x)-\frac12y^2(x)

Putting the value of y'(x) in equation (1)

y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))

Substituting x =0 and h= 0.2

y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]

\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]    [∵ y(0) =3 ]

\Rightarrow y(0.2)\approx 2.7

Substituting x =0.2 and h= 0.2

y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]

\Rightarrow y(0.4)\approx  2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]

\Rightarrow y(0.4)\approx 1.9926

Substituting x =0.4 and h= 0.2

y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]

\Rightarrow y(0.6)\approx  1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]

\Rightarrow y(0.6)\approx 1.6593

Substituting x =0.6 and h= 0.2

y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]

\Rightarrow y(0.8)\approx  1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]

\Rightarrow y(0.6)\approx 0.8800

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\Rightarrow y(1.0)\approx 0.9152

Therefore the value of y(1)= 0.9152.

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