All categories

3 years ago

How do you do question b?

Mathematics

1 answer:

notsponge [240]3 years ago

6 0

Part (b)

We use the result of part (a) and plug in (x,y) = (0,0). This is directly from the initial condition y(0) = 0.

$\arcsin(4y) = x^2 + C\\\\\arcsin(4*0) = (0)^2 + C\\\\\arcsin(0) = C\\\\0 = C\\\\C = 0\\\\$

-----------------

This means,

$\arcsin(4y) = x^2 + C\\\\\arcsin(4y) = x^2 + 0\\\\\arcsin(4y) = x^2\\\\4y = \sin(x^2)\\\\y = \frac{1}{4}\sin(x^2)\\\\$

is the solution with the initial condition y(0) = 0.

You might be interested in

If Naomi were to paint her living room alone, it would take 5 hours. Her sister Jackie could do the job in 8 hours. How many hou

Zielflug [23.3K]

Answer:

40/13

The decimal form is going to be 3.076

7 0

3 years ago

Forth grade math: What happens to the area of the triangle when one of the dimensions is doubled and halved?

Alja [10]

$\bf \textit{area of a triangle}\\\\ A=\cfrac{1}{2}bh~~ \begin{cases} b=base\\ h=height\\[-0.5em] \hrulefill\\ b=\stackrel{doubled}{2b} \end{cases}\implies A=\cfrac{1}{2}(2b)(h)\implies \stackrel{\textit{twice as the original area}}{A=bh} \\\\[-0.35em] ~\dotfill$

$\bf \textit{area of a triangle}\\\\ A=\cfrac{1}{2}bh~~ \begin{cases} b=base\\ h=height\\[-0.5em] \hrulefill\\ h=\stackrel{halved}{\frac{h}{2}} \end{cases}\implies A=\cfrac{1}{2}b\left( \cfrac{h}{2} \right)\implies \stackrel{\textit{half of the original area}}{A=\cfrac{1}{2}bh\left( \cfrac{1}{2} \right)}$

7 0

4 years ago

The completion time for an exam is normally distributed with expected value 75 minutes and variance 12 minutes. What is the prob

Debora [2.8K]

Answer:

14.89% or 0.1489

Step-by-step explanation:

First, find the z-score for both 70 and 80 minutes and their corresponding percentile.

$z= \frac{X - \mu }{SD} \\SD=\sqrt{V} =\sqrt{12}\\\mu =75\\z= \frac{X - 75}{\sqrt{12}}\\\\$

For X = 70 minutes:

$z= \frac{70 - 75}{\sqrt{12}}\\z=-1.4434$

This z-score is equivalent to the 7.445 th percentile, so the probability of a student finishing this exam in less than 70 minutes is 7.445%

or X = 80 minutes:

$z= \frac{80 - 75}{\sqrt{12}}\\z=1.4434$

This z-score is equivalent to the 92.555 th percentile, so the probability of a student finishing this exam in more than 80 minutes is 100- 92.555 = 7.445%

Therefore, the probability (P) of a student finishing this exam in less than 70 minutes or more than 80 minutes is:

P = 7.445% + 7.445% = 14.89%

5 0

3 years ago

One option in a roulette game is to bet $12 on red. (There are 18 red compartments, 18 black compartments, and two compartmen

Alona [7]

Answer:

The expected value is -$0,63

Step-by-step explanation:

The expected value of a discrete variable x is given by:

E(x)=x1*p(x1)+x2*p(x2)+...+xn*p(xn)

Where x1, x2, .. xn are the posibles results that the variable x can take and p(x1), p(x2),...,P(xn) are their respective probabilities.

In this case, we have two possible results if we bet $12 on red. This results are:

1. The balls lands on red: in this case we keep the $12 we paid and we awarded 12. That means, we gain $12 and the probability of this case is the division of the number of red compartments by total number of compartments. So:

x1=$12

p(x1)=18/38

2. The balls doesn't land on red: in this case we lose the $12 that we bet and the probability is the division of number of compartments that aren't red by the total number of compartments. So:

x2=-$12

p(x2)=20/38

Then, replacing x1, x2, p(x1) and p(x2) on the equation of the expected value, we get:

E(x)=($12)*(18/38) + (-$12)*(20/38)

E(x)=$-0.63

So, the expected value if we bet $12 on red is to lose $0.63

5 0

4 years ago

Find the HCF of 140,210,315

AVprozaik [17]

(On the far left, it says 140 at the top and then 5 to the left of 28)
The answer is 35, let me know if you can't read anything or have any questions. Hope this helped :)

3 0

3 years ago

Read 2 more answers