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Stels [109]
3 years ago
10

1.8×10^-2-3.9×10^-3 what is the answer

Mathematics
2 answers:
Alja [10]3 years ago
3 0
The answer is 0.0141
KonstantinChe [14]3 years ago
3 0

1.8×10^-2 = 0.018

3.9×10^-3 = 0.0039

So

1.8×10^-2-3.9×10^-3

= 0.018 - 0.0039

= -0.0141

<span>Answer in scientific form notation:</span>

<span>= -1.41 *10^-2</span>


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igomit [66]

Answer:

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

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)

3 0
3 years ago
What would be the first step to solve for n in the equation 3n-7=30?
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The first step would be to add 7 to both sides.
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4 0
2 years ago
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What is the square root of 64
iris [78.8K]

Answer:

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Step-by-step explanation:

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Mr. Thomas drove 75 miles in May. He drove 6 times as many miles in July as he did in May. He
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Answer:

1800 miles

Step-by-step explanation:

No. of miles driven by Mr. Thomas in May = 75

It is given that miles driven in July is 6 times of miles driven by Mr. Thomas in May(75 miles).

Thus

No. of miles driven by Mr. Thomas in July = 6 * No. of miles driven by Mr. Thomas in May   = 6*75 = 450 miles.

__________________________________________________

Another condition given   that miles driven in June is 4 times of miles driven by Mr. Thomas in July(450miles as calculated above).

Thus

No. of miles driven by Mr. Thomas in June = 4 * No. of miles driven by Mr. Thomas in July   = 4* 450 miles = 1800 miles.

No. of miles driven by Mr. Thomas in June is 1800 miles.

8 0
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