Ecd is a ___of abc cross line p

Mai runs around a 400 meter track at a constant speed of 250 meters per minute. How many minutes does it take Mai to complete 4

goldfiish [28.3K]

Answer:

6.4 minutes, or 6 minutes and 24 seconds

Step-by-step explanation:

She is running 4 laps, so she is running 1600 meters total. If she runs 250 meters per minute, divide 1600 by 250 to determine how many minutes it will take...

1600/250 = 6.4

Which is 6 minutes + 0.4 minutes

*there are 60 seconds in a minute, so the 0.4 represents 40% of another minute. Multiply 60 by 0.4 to see how many seconds this is...

(0.4)60 = 24 seconds,

So she ran the laps in 6 minutes and 24 seconds

7 0

3 years ago

Read 2 more answers

Solve for x.* (4x + 7)° (6x + 3)° Sign

Whitepunk [10]

24 x ^2 + 54 x + 21,

if you would like me to explain how to do it, I can in the comments.

3 0

3 years ago

Which is the correct answer. Quickly help please

Darya [45]

I think is the third one

3 0

3 years ago

2.4 x 0.87 help help help

yawa3891 [41]

Answer:

2.088

Step-by-step explanation:

Trust me I am in high school honors algebra with almost an average of 100.

5 0

3 years ago

Read 2 more answers

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