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3 years ago

What is 18 divided by 2/3?

2 answers:

6 0

Division is basically
x divided by y=x/y so
18 divided by 2/3= $\frac{18}{ \frac{2}{3} }$
we want the bottom number to be =1 so we multiply 2/3 by 3/2 to make it 1 (or 6/6=1), but we have to multiply the top number by 3/2 also so that we don't change the fraction and make our work useless so

$\frac{18}{ \frac{2}{3} } times \frac{ \frac{3}{2} }{ \frac{3}{2} } = \frac{ \frac{54}{2} }{ \frac{6}{6} }= \frac{27}{1}=27$

the answe ris 27

harina [27]3 years ago

4 0

18 divided by 2/3 is 27. 2/3 is .66666. Once you do 18/.6666, you obtain 27 as your answer.

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Which function , A or B has a greater rate of change ? Be sure to include the values for the rates of change in your answer . Ex

arsen [322]

<h2>Answer:</h2>

A rate of change tells us how one quantity changes in relation to another quantity. In a mathematical language, we can write this as follows:

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By taking two points we can get the rate of change, so let's take $(1,5) \ and \ (2,7)$ :

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Let's take $(1,1) \ and \ (2,4)$ :

$ROC=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \\ \\ ROC=\frac{4-1}{2-1} \\ \\ ROC=\frac{3}{1} \\ \\ \boxed{ROC=3}$

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3 years ago

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Step-by-step explanation:

4 0

3 years ago

How do I find the integral<br> ∫10(x−1)(x2+9)dxint10/((x-1)(x^2+9))dx ?

defon

$\int\frac{10}{(x-1)(x^2+9)}\ dx=(*)\\\\\frac{10}{(x-1)(x^2+9)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+9}=\frac{A(x^2+9)+(Bx+C)(x-1)}{(x-1)(x^2+9)}\\\\=\frac{Ax^2+9A+Bx^2-Bx+Cx-C}{(x-1)(x^2+9)}=\frac{(A+B)x^2+(-B+C)x+(9A-C)}{(x-1)(x^2+9)}\\\Updownarrow\\10=(A+B)x^2+(-B+C)x+(9A-C)\\\Updownarrow\\A+B=0\ and\ -B+C=0\ and\ 9A-C=10\\A=-B\ and\ C=B\to9(-B)-B=10\to-10B=10\to B=-1\\A=-(-1)=1\ and\ C=-1$

$(*)=\int\left(\frac{1}{x-1}+\frac{-x-1}{x^2+9}\right)\ dx=\int\left(\frac{1}{x-1}-\frac{x+1}{x^2+9}\right)\ dx\\\\=\int\frac{1}{x-1}\ dx-\int\frac{x+1}{x^2+9}\ dx=\int\frac{1}{x-1}-\int\frac{x}{x^2+9}\ dx-\int\frac{1}{x^2+9}\ dx=(**)\\\\\#1\ \int\frac{1}{x-1}\ dx\Rightarrow\left|\begin{array}{ccc}x-1=t\\dx=dt\end{array}\right|\Rightarrow\int\frac{1}{t}\ dt=lnt+C_1=ln(x-1)+C_1$

$\#2\ \int\frac{x}{x^2+9}\ dx\Rightarrow \left|\begin{array}{ccc}x^2+9=u\\2x\ dx=du\\x\ dx=\frac{1}{2}\ du\end{array}\right|\Rightarrow\int\left(\frac{1}{2}\cdot\frac{1}{u}\right)\ du=\frac{1}{2}\int\frac{1}{u}\ du\\\\\\=\frac{1}{2}ln(u)+C_2=\frac{1}{2}ln(x^2+9)+C_2$

$\#3\ \int\frac{1}{x^2+9}\ dx=\int\frac{1}{x^2+3^2}\ dx=\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C_3\\\\therefore:\\\\\#1;\ \#2;\ \#3\Rightarrow(**)=ln(x-1)+C_1-\frac{1}{2}ln(x^2+9)+C_2-\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C_3$

$\boxed{=ln(x-1)-\frac{1}{2}ln(x^2+9)-\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C}$

4 0

4 years ago