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2 years ago

Need to find the value of y

Mathematics

1 answer:

alexgriva [62]2 years ago

3 0

Answer:

√55

Step-by-step explanation:

Notice that the small triangle in the bottom corner shares an angle with the overall triangle. Also, they are both right triangles. Therefore, they are similar triangles.

Notice that the large triangle at the top shares an angle with the overall triangle and is also a right triangle. Therefore it is also similar to the overall triangle and the smaller triangle.

Writing a proportion between the small and large triangles:

y / 11 = 5 / y

y² = 55

y = √55

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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)

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

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

True or false 120% of a whole number is always greater than the number? and why?

Stells [14]

I can only use math to explain it. What is 100%?
100% = $\frac{100}{100}$
percent means per 100.
so 120% = 120/100 = 1.2
120% of a whole number. For example 120% * 10 = $\frac{120}{100} * 10$ = 12
is it greater than 10.

5 0

2 years ago

Read 2 more answers

Find the area of the figure shown below.<br> 10 cm<br> 5 cm<br> 17 cm

stira [4]

Answer:

where is the figure?

Step-by-step explanation:

I can't see a figure

8 0

2 years ago

Read 2 more answers

-4<-2(3x+9)+5<1 Please help

Yanka [14]

Answer: if ur looking for interval notation its (-7/3,-3/2) ; if ur solving for x then its -7/3<x<-3/2

Step-by-step explanation:

5 0

3 years ago

Could you please look at this question?

MrMuchimi

Answer:

a. see attached

b. H(t) = 12 -10cos(πt/10)

c. H(16) ≈ 8.91 m

Step-by-step explanation:

The cosine function has its extreme (positive) value when its argument is 0, so we like to use that function for circular motion problems that have an extreme value at t=0. The midline of the function needs to be adjusted upward from 0 to a value that is 2 m more than the 10 m radius. The amplitude of the function will be the 10 m radius. The period of the function is 20 seconds, so the cosine function will be scaled so that one full period is completed at t=20. That is, the argument of the cosine will be 2π(t/20) = πt/10.

The function describing the height will be ...

H(t) = 12 -10cos(πt/10)

The graph of it is attached.

See part a.

The wheel will reach the top of its travel after 1/2 of its period, or t=10. Then 6 seconds later is t=16.

H(16) = 12 -10cos(π(16/10) = 12 -10cos(1.6π) ≈ 12 -10(0.309017) ≈ 8.90983

The height of the rider 6 seconds after passing the top will be about 8.91 m.

8 0

2 years ago