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masya89 [10]
3 years ago
15

A motorboat can go 16 miles downstream on a river in 20 minutes. It takes 30 minutes for this boat to go back upstream the same

16 miles. Let x = the speed of the boat. Let y = the speed of the current. a. Write an equation for the motion of the motorboat downstream. b. Write an equation for the motion of the motorboat upstream. c. Find the speed of the current.
Mathematics
1 answer:
Katyanochek1 [597]3 years ago
7 0

Answer:

A) d = 48t, d in miles, t in hours

B)d = 32t, d in miles, t in hours

C)y = 8 mph

Step-by-step explanation:

16 miles in 20 mins = 48 mph

16 miles in 30 mins = 32 mph

The boat speed is the average = (48 + 32)/2 = 40 mph

The current is the difference = 8 mph

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7cm

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Answer:

Multiply the a nagative to what is inside the paranthesis on the right side of the nagative sign then combine like variables. This will give you your answer.

Step-by-step explanation:

Multiply the a nagative to what is inside the paranthesis on the right side of the nagative sign then combine like variables. This will give you your answer.

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3 years ago
Match the lengths of the hypotenuse and one leg of a right triangle to the length of the other leg.
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1. D
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3 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
3 years ago
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Ray Of Light [21]

Answer:

y - 8 = 3(x - 8)

Step-by-step explanation:

Point-Slope Form: y + y₁ = m(x + x₁)

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Step 1: Define

Point (-8, -8)

Slope <em>m</em> = 3

Step 2: Write in point-slope form

y - 8 = 3(x - 8)

6 0
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