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3 years ago

-10y<60 (Please solve and show work)

2 answers:

7 0

Y can realistically be any positive number, since a negative number multiplied by a positive number's product is negative. a negative number will always be less than a positive number, in this case the positive number being 60.

<span>a negative number multiplied by another negative number's product is positive, so as long as the negative numbers are less than -6 (because -10 times -6 is 60, and 60 is not less than 60) y can be negative.</span>

I hope this helps!

LekaFEV [45]3 years ago

3 0

When you divide or multiply by a negative, flip the sign
divide both sides by -10
flip sign

y>-6

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Find the domain and range of the following function f(x) = 31x + 71 - 2?

Harman [31]

Answer: domain and range are all real numbers!

Step-by-step explanation:

4 0

3 years ago

Consider the ODE, dy dx = y 2 1 + x (2) subject to condition y = 1 when x = 0, use your Euler code from class (modified if neces

makkiz [27]

Answer:

Computation.

Step-by-step explanation:

I'm not really sure if that's the analytical solution of the inital value problem,

because y(0)=11-ln(1-0)(3)=11. Howevwer, let us procede with the given values...

Let us assume that we are going to use euler with n=2 (two steps) and h=0.2(the size of each step)

The update rules of the Euler Methode are

$X_i = X_{i-1}+h=X_0+ih$

$m_i=\dfrac{dY}{dX} \biggr \rvert_{x_i} \\\\Y_{i+1}=m_i\cdot h+y_i$

Since the initial value problem tells us that Y=1 when X=0, we know that

$X_0=0$ and that $Y_0=1$ . Then, we have

$X_0=0\\\\X_1=0.2\\\\X_2=0.4$

and

$Y_0=1\\\\m_0=21 \cdot Y_0 + X_0 \cdot 2=21 \cdot 1 + 0=21\\\\Y_1=0.2 \cdot 21 + 1 =5.2\\\\m_1=21 \cdot Y_1 + X_1\cdot 2=21 \cdot 5.2 + 0.2 \cdot 2=109.6\\ \\Y_2=m_1 \cdot h + Y_1 = 109.6 \cdot 0.2 + 5.2= 27.12$

which gives us the points (0,1), (0.2, 5.2) and (0.4, 27.12).

Now, since we want to compare the analyticaland the Euler result, we first compute the value of y=11-ln(1-x)(3) for the values x=0, 0.2 and 0.4. We get that

$y(0)=11-\ln(1-0)(3)=11-ln(1)(3)=11\\\\Y(1)=11-\ln(1-0.2)(3)=11.67\\\\Y(2)=11-\ln(1-0.4)(3)=12.53$

and we compute $Y(i)-Y_i$ for each i.

It holds

$Y(0)-Y_0=11-1=10\\\\Y(1)-Y_1=11.67-5.2=6.47\\\\Y(2)-Y_2=12.53-27.12=-14.59$

which tells us that we have a really bad approximation, as I already stated there must be a mistake in the analytical solution since the intial values don't coincide. Also note that the curve that we get using the euler methose is growing faster than the analitical solution.

4 0

3 years ago

18-3+5-12÷4. I am needing help on how to work this problem

AleksandrR [38]

Answer

Step-by-step explanation:

12 divided by 4=3

18-3+5-3

5 minus 3 = 2

18-3+2

18 minus 3 = 15

15+2= 17