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3 years ago

Find the general solution of the differential equation. (use c for the constant of integration.) dr dt = 8et 1 − e2t

1 answer:

5 0

Try this solution:
if given dr/dt=8e^t -e^(2t), then
$\frac{dr}{dt}=8e^t-e^{2t}; \ \int\ {dr} \, =\int\ (8e^t-e^{2t})dt; \ r=8e^t- \frac{1}{2}e^{2t}+C.$

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Is there a smallest decimal that is greater than the number 2? Explain your answer.

olga nikolaevna [1]

Answer:

Technically, yes. It's called epsilon, which is defined as an infinitely small number. So

2 + epsilon is the smallest number greater than 2. But for practical purposes no there isn't.

Step-by-step explanation:Not without limits. You can always move the .1 one place further from the interring. For example,

2.1>2.01

and

2.01>2.001

So, unless there are a limited number of decimal spaces, you can continually add an infinite amount of zeros behind the decimal point, followed by a one.

If you use two or three decimal spaces as a standard in your class, then the smallest decimal greater than 2 would be 2.01 or 2.001, respectively.

8 0

3 years ago

Can someone please help

EastWind [94]

You basically just switch it around and subtract the fractions

7 0

3 years ago

Please solve the following quadratic equation and show your work: x^2 -x =30

saveliy_v [14]

Rewriting the equation as a quadratic equation equal to zero:
x^2 - x - 30 = 0
We need two numbers whose sum is -1 and whose product is -30. In this case, it would have to be 5 and -6. Therefore we can also write our equation in the factored form
(x + 5)(x - 6) = 0
Now we have a product of two expressions that is equal to zero, which means any x value that makes either (x + 5) or (x - 6) zero will make their product zero.
x + 5 = 0 => x = -5
x - 6 = 0 => x = 6
Therefore, our solutions are x = -5 and x = 6.

8 0

3 years ago

2x 1/2 x3 2/3 im goin to be need a lot of helpppp

RSB [31]

Answer:

3.66666666667

3.67(rounded to the nearest tenth)

Step-by-step explanation:

2x 1/2 x3 2/3 = 3.66666666667

<em>Ace Carlos</em>

<em>Developer</em>

4 0

4 years ago

Read 2 more answers

Write the quadratic function with the following transformations: vertical shrink by 3, left 4, down 9.

xz_007 [3.2K]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\
% left side templates
\begin{array}{llll}
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}\\\\
--------------------\\\\$

$\bf \bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}
\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\$

$\bf \bullet \textit{ vertical shift by }{{ D}}\\
\left. \qquad \right. if\ {{ D}}\textit{ is negative, downwards}\\\\
\left. \qquad \right. if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}$

with that template in mind, let's see

so, the parent equation could be
y = x² ====> y = A(x + C)² + D

shrink by 3, A = 3
left 4, C = +4
down 9, D = 9

7 0

4 years ago