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Keith_Richards [23]
3 years ago
11

Your classmate is starting a new fitness program. He is planning to ride his bicycle 30 minutes every day. He burns 7 calories p

er minute bicycling at 11 mph and 10.75 calories per minute bicycling at 15 mph. How long should he bicycle at each speed to burn 300 ​calories?
Mathematics
1 answer:
AlexFokin [52]3 years ago
4 0

Set your equations up algebraically.

At 11mph: 43 minutes to burn 300 calories

At 15mph: 28 minutes to burn 300 calories

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Hard math question!
luda_lava [24]
First, let me show you some notation.

To show a matrix is an inverse of another matrix, we write A^{-1}

-1 is not an exponent. It just shows that a matrix is an inverse of another matrix.

For a 2x2 matrix, we can get the inverse by first making b and c negatives and swap the positions of a and d.

Then multiply each entry in the matrix by 1 divided by the determinant.

\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]^{-1} = 
  \frac{1}{ad - bc}\left[\begin{array}{ccc}d&{-b}\\{-c}&a\end{array}\right] =  \\  \\ \\ \left[\begin{array}{ccc}d(\frac{1}{ad-bc})&{-b}(\frac{1}{ad-bc}) \\ {-c}(\frac{1}{ad-bc}) &a(\frac{1}{ad-bc}) \end{array}\right]

I hope this helped!
8 0
3 years ago
Plz use Differential equation method to solve this:
Oduvanchick [21]

Answer:

Step-by-step explanation:

We have the differential equation y' = y + \frac{x}{y} with initial conditions y(0)=1.

First, notice that the equation can be rewritten as

y'-y =xy^{-1},

which is a Bernoulli equation. Once we have recognized the type of the equation we know how to continue. Recall that a Bernoulli equation has the general form

y'+p(x)y=q(x)y^n.

In this particular case we have n=-1. This kind of equation is solved by the change of variable z=y^{1-n}. In our exercise we get z=y^{1-(-1)}=y^2. Now we take derivatives and get

z'=2yy' which es equivalent to \frac{z'}{2y}=y'.

Then, we substitute the value of y' we have obtained in the original equation:

\frac{z'}{2y}-y = xy^{-1}.

The next step is to multiply the whole equation by 2y, in order to eliminate the denominator of z'. Thus,

z' -2y^2=2x.

Recall that y^2=z, then

z' -2z=2x.

This last equation is a linear equation, which has general solution

z(x) = \exp\left(-\int(-2)dx\right)\left(\int 2x \exp\left(\int(-2)dx + C\right)\right).

So, let us calculate the integral that appear in the formula:

\int(-2)dx = -2x

\int 2x e^{-2x}dx = -\frac{\left(2x+1\right)e^{-2x}}{2}.

Then, the solution for z is

z(x) = e^{2x}\left(-\frac{\left(2x+1\right)e^{-2x}}{2} + C\right) = -\frac{\left(2x+1\right)}{2} + Ce^{2x}.

Now, we return the change of variable:

(y(x))^2 =-\frac{\left(2x+1\right)}{2} + Ce^{2x}.

The last step is to find the value of the constant C. In order to do this, substitute the initial value:

(y(0))^2 = 1 =\frac{\left(2\cdot 0+1\right)}{2} + Ce^{2\cdot 0} = -\frac{1}{2} + C.

Thus, we have the equation

1=-\frac{1}{2} + C that gives C=\frac{3}{2}.

Therefore,

(y(x))^2 = -\frac{\left(2x+1\right)}{2} + \frac{3}{2}e^{2x}.

8 0
4 years ago
Find measure of Arc BC
iren2701 [21]

Answer:

arc BC = 140

Step-by-step explanation:

Comment

Pick a point inside the circle. It should look like this point is the center of the circle. Call this point D. Draw lines CD and BD

CD makes a right angle with AC

BD makes a right angle with AB

The central angle formed by The given information above is

<BDC = 360 - 90 - 90 - 40

<BDC = 140.

Answer The arc measurement of BC = 140

5 0
2 years ago
Enter the equation of the line in slope-intercept form. Slope is − 1 2 , and (−9, 2) is on the line. The equation of the line is
Rashid [163]

Answer:

y=-12-2

Step-by-step explanation:

y=mx-b

m=slope

slope =-12

b= y intercept

y intercept= 2

    y=-12-2

3 0
4 years ago
HELP PLEASE ASAP NO LINKS OR SCAMS ​
kozerog [31]

Answer:

1

   a. 5

   b. 32

   c.  130

   d. 26

2

    a. 5.65234

    b. 5652.34

    c. 8.1253758

    d. 8125375.8

Step-by-step explanation:

3 0
3 years ago
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