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2 years ago

Can you figure out the missing number

Mathematics

2 answers:

astra-53 [7]2 years ago

8 0

Answer:

It is definitely 6

Step-by-step explanation:

-_-

gayaneshka [121]2 years ago

6 0

Answer:

Im not sure exactly what this is but it seems the answer is 6 because the only number 1-9 thats not there!

Step-by-step explanation:

Hope this helps!! :)

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Geometry First Semester.<br>Can someone please help with this?

Len [333]

1-a
2-d
3-a
4-b
5-d
6-b
7-a
8-c
9- i don't know the answer, just guess....
10 - d

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3 years ago

Which sum of difference is equivalent to the following expression? 5-3x/5

Aleks04 [339]

Answer:

1-3x/5

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Step-by-step explanation:

7 0

2 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

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

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

$\Rightarrow y(0.6)\approx 0.8800$

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

Which of the fractions shown in the box below is<br> the largest?<br> 3/4, 5/6, 11/12, 19/24

ruslelena [56]

Answer:

24 is the largest fraction fell by 12 to 19

6 0

2 years ago

In 16 years, Ben will be 3 times as old as he is now. How old is he?

Igoryamba

First, do 16 x 3 to get 48. Then subtract 16 and youll get 32

3 0

3 years ago

Read 2 more answers

Can you figure out the missing number​

Can you figure out the missing number