A woman works out by running and swimming. When she runs, she burns 7 calories per minute. When she swims, she burns 8 calories per minute. She wants to burn at least 336 calories in her workout. Write an inequality that describes the situation. Let x represent the number of minutes running and y the number of minutes swimming. Because x and y must be positive, limit the boarders to quadrant I only.

Question

A woman works out by running and swimming. When she runs, she burns 7 calories per minute. When she swims, she burns 8 calories per minute. She wants to burn at least 336 calories in her workout. Write an inequality that describes the situation. Let x represent the number of minutes running and y the number of minutes swimming. Because x and y must be positive, limit the boarders to quadrant I only.

8x+7y>=336

Sn: 1^2+...+(3n-2)^2 = n(6n^2-3n-1)/2
   S1: 1^2 = 1(6*1^2-3(1)-1)/2
 S2: 1^2+4^2 = 2(6*2^2-3*2-1)/2
 S3: 1^2+4^2+7^2 = 3(6*3^2-3*3-1)/2

There all true but I don't know the induction for this

Answer 1

I do not have time

NOTE
in your previous question, the determinant of the 3 by 3 coef matric is zero, can not solve.

Answer 2

Oh okay thank you so much

Answer 3

Your inequality is correct. I have no idea what meaning of the rest is.

Think on a graph x,y, quadrant 1.

all the area inside the triangle boardered by x axis, y axis, and the line 8x+7y=335, the region inside is not an allowable solution. On the line, and outside the triangle, any point is allowed and is a solution. Thus, 2000,4000 is an allowable solution, whereas 2,5 is not.

Answer 4

Thank you!

Answer 5

To understand the induction for the given series, let's break it down step by step.

We have the series Sn: 1^2 + 4^2 + ... + (3n-2)^2. Our goal is to prove that this series can be expressed as n(6n^2-3n-1)/2.

First, we will show that the formula holds for the base case n=1:
Sn: 1^2 = 1(6*1^2-3*1-1)/2
1 = 1(6-3-1)/2
1 = 1(2)/2
1 = 1

Since the formula is true for n=1, we move on to the second step of induction, which is to assume that the formula holds for a certain value of n, and then prove it holds for the next value (n+1).

Assuming the formula holds for some value k:
Sk: 1^2 + 4^2 + ... + (3k-2)^2 = k(6k^2-3k-1)/2

Now, let's prove that the formula holds for the next value (k+1):
S(k+1): 1^2 + 4^2 + ... + (3(k+1)-2)^2 = (k+1)(6(k+1)^2-3(k+1)-1)/2

Expanding the equation on the left side:
S(k+1): S(k) + (3(k+1)-2)^2

Replacing S(k) with the assumed formula:
S(k+1): k(6k^2-3k-1)/2 + (3(k+1)-2)^2

Simplifying the equation on the right side:
S(k+1): k(6k^2-3k-1)/2 + (3k+1)^2

Expanding and simplifying further:
S(k+1): (6k^3 - 3k^2 - k)/2 + (9k^2 + 6k + 1)

Combining like terms:
S(k+1): (6k^3 - 3k^2 + 9k^2 - k + 6k + 1)/2

Simplifying:
S(k+1): (6k^3 + 6k^2 + 5k + 1)/2

Factoring out a common factor of 6:
S(k+1): 6(k^3 + k^2 + 5k)/2 + 1/2

Simplifying:
S(k+1): 3(k^3 + k^2 + 5k) + 1/2

Factoring out a common factor of k:
S(k+1): 3k(k^2 + k + 5) + 1/2

Now we can see that the right side of the equation matches the form of the formula we assumed, but with n replaced by (k+1):
S(k+1): (k+1)(6(k+1)^2-3(k+1)-1)/2

Therefore, we have proved that if the formula is true for some value k, then it is also true for the next value (k+1).

By completing the base case and the induction step, we have proved that the formula n(6n^2-3n-1)/2 represents the given series Sn: 1^2 + 4^2 + ... + (3n-2)^2.