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Verdich [7]
3 years ago
12

If your heart beats 65 times per minute. How many times does the heart beat in a day, a year, a lifetime?

Mathematics
2 answers:
Kobotan [32]3 years ago
6 0
give me a dollar manuuhuuuhuuuyy
tekilochka [14]3 years ago
5 0
65/minute
3900/hour
93600/day
34164000/year
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Plzz solve this easy calculus :')
Digiron [165]

Answer:

\sin \: x -  \cos  x \:  + c

Step-by-step explanation:

\int \frac{cos2x}{ \sqrt{1 - sin2x} } dx \\  \\  =  \int \frac{ {cos}^{2}x -  {sin}^{2}x }  { \sqrt{ {sin}^{2}x +  {cos}^{2}x - 2sinx \: cosx  }  } dx \\  \\ =  \int \frac{ ({cos}x -  {sin}x )({cos}x  +   {sin}x )}  { \sqrt{  {({cos}x -  {sin}x )}^{2}  }  } dx \\  \\ =  \int \frac{ ({cos}x -  {sin}x )({cos}x  +   {sin}x )}  {{  {({cos}x -  {sin}x )} }  } dx \\  \\  =  \int({cos}x  +   {sin}x)dx \\  \\  = \int{cos}x  \: dx +    \int{sin}x \: dx \\  \\  =  \sin \: x -  \cos  x \:  + c

3 0
2 years ago
What are the coordinates of the endpoints of the midsegment for DEF that is parallel to DE
Nutka1998 [239]

Answer:

\left(\dfrac{x_D+x_F}{2},\dfrac{y_D+y_F}{2}\right),  \left(\dfrac{x_E+x_F}{2},\dfrac{y_E+y_F}{2}\right).

Step-by-step explanation:

Let points D, E and F have coordinates (x_D,y_D),\ (x_E,y_E) and (x_F,y_F).

1. Midpoint M of segment DF has coordinates

\left(\dfrac{x_D+x_F}{2},\dfrac{y_D+y_F}{2}\right).

2. Midpoint N of segment EF has coordinates

\left(\dfrac{x_E+x_F}{2},\dfrac{y_E+y_F}{2}\right).

3. By the triangle midline theorem, midline MN is parallel to the side DE of the triangle DEF, then points M and N are endpoints of the midsegment for DEF that is parallel to DE.

6 0
3 years ago
Find the area of a rhombus with diagonals 9 ft. and 12 ft
aleksklad [387]
The answer would be 54, because A= PQ over 2 = 9 x 12 over 2 = 54
3 0
3 years ago
- 4j - 1 - 4j+6<br><br> Write that expression in the simplest form
Anika [276]

Answer:

-8j+5

Step-by-step explanation:

the answer is -8j+5

4 0
2 years ago
Solve the 3 × 3 system shown below. Enter the values of x, y, and z. X + 2y – z = –3 (1) 2x – y + z = 5 (2) x – y + z = 4 (3)
Svet_ta [14]
<h2>Answer:</h2>

\boxed{x=1, \y=-1, \ z=2}

<h2>Step-by-step explanation:</h2>

We will use the Gaussian elimination method to solve this problem. To do so, let's follow the following steps:

Step 1: Let's multiply first equation by −2. Next, add the result to the second equation. So:

\begin{array}{ cccc }~~ x&+~~2~ y&-~~~~~ z&~=~-3\\&-~~~5~ y&+~~3~ z&~=~11\\~~ x&-~~~~~ y&+~~~~ z&~=~4\end{array}

Step 2: Let's multiply first equation by −1. Next, add the result to the third equation. Thus:

\begin{array}{ cccc }~~ x&+~~2~ y&-~~~~~ z&~=~-3\\&-~~~5~ y&+~~3~ z&~=~11\\&-~~~3~ y&+~~2~ z&~=~7\end{array}

Step 3: Let's multiply second equation by −35, Next, add the result to the third equation. Therefore:

\begin{array}{ cccc }~~ x&+~~2~ y&-~~~~~ z&~=~-3\\&-~~~5~ y&+~~3~ z&~=~11\\&&+~~\frac{ 1 }{ 5 }~ z&~=~\frac{ 2 }{ 5 }\end{array}

Step 4: solve for z, then for y, then for x:

\frac{ 1 }{ 5 } ~ z & = \frac{ 2 }{ 5 } \\ \\ \boxed{z & = 2}

-5y+3z &= 11\\-5y+3\cdot 2 &= 11\\ \\ \boxed{y &= -1}

By substituting y=-1 \ and \ z=2 into the first equation, we get the x. So:

x+2(-1)-2=-3 \\ \\ x-2-2=-3 \\ \\ \boxed{x=1}

6 0
3 years ago
Read 2 more answers
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