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2 years ago

10

I need to find the area

1 answer:

SCORPION-xisa [38]2 years ago

4 0

Answer: 24 sq. ft

Step-by-step explanation: 20 x 36 - (36 - 20 / 2 ) x (20 - 14 )= 24 sq. ft

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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

Determine if graph is a function

Zepler [3.9K]

No it is not a function.

8 0

3 years ago

The amounts of food waste generated in a region during two years were 37,500,000 tons and 30,400,000 tons. What was the total fo

alex41 [277]

Answer:

6.79 ⋅ 107

Step-by-step explanation:

3 0

2 years ago

Please answer this - i'm so desperate!

Ksenya-84 [330]

Try 8, 7, 4, 1 (I believe that’s the right answer but I’m not sure)

7 0

3 years ago

The value of x in the equation log(x-1)9=2 is

mel-nik [20]

$\log_{(x-1)} 9=2$

the domain:
$x-1 >0 \ \land \ x-1 \not=1 \\
x>1 \ \land \ x \not= 2 \\
x \in (1; 2) \cup (2;+\infty)$

the equation:
$\log_{(x-1)}9=2 \\
(x-1)^2=9 \\
\sqrt{(x-1)^2}=\sqrt{9} \\
|x-1|=3 \\
x-1=3 \ \lor \ x-1=-3 \\
x=4 \ \lor \ x=-2$

4 is in the domain
-2 is not in the domain

The answer:
$x=4$

***
Also, the answer in 13 is {8}. -3 is not in the domain. Replace x with -3 and you'll see:
$x=\sqrt{5x+24} \\
-3=\sqrt{5 \times (-3)+24} \\
-3=\sqrt{-15+24} \\
-3=\sqrt{9} \\
-3=3$
It's not true so -3 isn't a solution to this equation.

4 0

3 years ago

Read 2 more answers