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sladkih [1.3K]
2 years ago
7

kevin completed 10 problems of his homework during lunch this is 2/7 of the total assignment how many problem was Kevin assigned

for homework
Mathematics
1 answer:
Maurinko [17]2 years ago
6 0

Answer:

35 problems

Step-by-step explanation:

We know that 10 problems are 2/7 of his total assignment, so we can set up the following equation,

\frac{10}{x\\} = \frac{2}{7}

(side note: x is the denominator because we don't know how many problems it is out of yet)

We can cross multiply which gives us: (7x10) and 2(x)

70 = 2x (set them equal)

x = 35 (divide 70 by 2)

Hope this helps!

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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)

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Substituting x =0 and h= 0.2

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

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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]

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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]

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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

Substituting x =0.8 and h= 0.2

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

\Rightarrow y(1.0)\approx  0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]

\Rightarrow y(1.0)\approx 0.9152

Therefore the value of y(1)= 0.9152.

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