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steposvetlana [31]

2 years ago

13

Please help me I'm a bit stuck on this problem.

1 answer:

bixtya [17]2 years ago

7 0

5 times when all 5 times are up its 7:30 which if 3/4 of a hour which is 45 and is that time4learning?

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. Consider the leading term of the polynomial function. What is the end behavior of the graph? Describe the end behavior and pro

QveST [7]

Because this is a fifth degree polynomial, it's an odd function. Odd functions have one tail go up and the other down. Because the leading coefficient is a negative number, 3 to be exact, the left one goes up and the right goes down.

3 0

2 years ago

What is the difference in length between a 1 1/4 inch button and a 3/8 inch button?

charle [14.2K]

1 1/4= 1*4+1 /4. Or 5/4

In order to subtract you must have same denominator. 5/4 -3/8

Multiply 5/4 times 2/2. =. 10/8. -3/8

The difference is 7/8

8 0

2 years ago

Who is good at the maths/ algebra??

Goryan [66]

8 < 2b
8/2 < b
4 < b

2b + 15 < 5
2b < 5 - 15
2b < -10
b < -10/2
b < -5

answer is : b > 4 or b < -5

3 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

2 years ago

Question in screenshot.

Kitty [74]

Your answer is b. hope this helps :)

4 0

3 years ago