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STatiana [176]
3 years ago
11

1058 rounded to the nearest hundred

Mathematics
1 answer:
GalinKa [24]3 years ago
7 0

Answer:

1100

Step-by-step explanation:

Rounded to nearest hundred

Here we will tell you what 1058 is rounded to the nearest hundred and also show you what rules we used to get to the answer.

Remember, we did not necessarily round up or down, but to the hundred that is nearest to 1058.

First, 1058 rounded to the nearest hundred is:

1100

When rounding to the nearest hundred, like we did with 1058 above, we use the following rules:

A) We round the number up to the nearest hundred if the last two digits in the number are 50 or above.

B) We round the number down to the nearest hundred if the last two digits in the number are 49 or below.

C) If the last two digits are 00, then we do not have to do any rounding, because it is already to the hundred.

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x = c1 cos(t) + c2 sin(t) is a two-parameter family of solutions of the second-order DE x'' + x = 0. Find a solution of the seco
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Answer:

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Step-by-step explanation:

The problem is very simple, since they give us the solution from the start. However I will show you how they came to that solution:

A differential equation of the form:

a_n y^n +a_n_-_1y^{n-1}+...+a_1y'+a_oy=0

Will have a characteristic equation of the form:

a_n r^n +a_n_-_1r^{n-1}+...+a_1r+a_o=0

Where solutions r_1,r_2...,r_n are the roots from which the general solution can be found.

For real roots the solution is given by:

y(t)=c_1e^{r_1t} +c_2e^{r_2t}

For real repeated roots the solution is given by:

y(t)=c_1e^{rt} +c_2te^{rt}

For complex roots the solution is given by:

y(t)=c_1e^{\lambda t} cos(\mu t)+c_2e^{\lambda t} sin(\mu t)

Where:

r_1_,_2=\lambda \pm \mu i

Let's find the solution for x''+x=0 using the previous information:

The characteristic equation is:

r^{2} +1=0

So, the roots are given by:

r_1_,_2=0\pm \sqrt{-1} =\pm i

Therefore, the solution is:

x(t)=c_1cos(t)+c_2sin(t)

As you can see, is the same solution provided by the problem.

Moving on, let's find the derivative of x(t) in order to find the constants c_1 and c_2:

x'(t)=-c_1sin(t)+c_2cos(t)

Evaluating the initial conditions:

x(0)=-1\\\\-1=c_1cos(0)+c_2sin(0)\\\\-1=c_1

And

x'(0)=2\\\\2=-c_1sin(0)+c_2cos(0)\\\\2=c_2

Now we have found the value of the constants, the solution of the second-order IVP is:

x=-cos(t)+2sin(t)

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Answer:

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