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Effectus [21]
3 years ago
6

* Worth 90 points *

Mathematics
2 answers:
Tresset [83]3 years ago
7 0

Answer:

no question

Step-by-step explanation:

ioda3 years ago
3 0

Answer:

there is no question

Step-by-step explanation:

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Bria ordered 145 centimeters of fabric. Jayleen ordered 1.5 Meters or fabric. who ordered more fabric
liq [111]
Jayleen ordered more fabric by 5 centimeters...
Bria ordered 145 cm and Jayleen had 1.5 m ( 150 cm ) 
So 145<150 
Therefore Jayleen ordered more than Bria.
4 0
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Anyone know? Please help ASAP! Thx!
sdas [7]
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8 0
3 years ago
If the area is 1808.64 what is the circumference
Dafna1 [17]
Area of circle = πr²

Find radius:
πr² = 1808.64
r² = 1808.64 ÷ π
r² = 575.71
r = √575.71
r = 23.99 units

Find Circumference:
Circumference = 2πr
Circumference = 2π (23.99)
Circumference = 150.76 units

Answer: Circumference = 150.76 units
3 0
4 years ago
What does 7×42 equal
miss Akunina [59]
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5 0
3 years ago
Read 2 more answers
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

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.

4 0
3 years ago
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