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AlekseyPX
3 years ago
10

Joanna ran a mile in physical education class. After resting for one hour, her heart rate was 60 beats per minute. If her heart

rate decreased by 2/5, what was her heart rate immediately after she ran the mile?
Mathematics
1 answer:
Ugo [173]3 years ago
4 0
So, if 60 beats was decreased by 2/5 before, we know that 60 beats = 3/5 of her heart rate after she ran the mile.

(Because 1 - 2/5 = 3/5)

So, 3/5 = 60, and we're trying to find 1.

So I would first divide both sides by 3.

3/5 = 60
/3      /3

1/5 = 20

Then multiply both sides by 5.

1/5 * 5 = 20 * 5

1 = 100

So her heart rate was 100 beats per minute immediately after she ran the mile.
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Read 2 more answers
Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval. ex = 3 − 2x, (0
laila [671]

Answer:

f(x) is continuous on the interval [0, 1], f(0) = <u>-2</u> , and f(1) = <u>1.718</u> . Since f(0) < 0 < f(1) , there is a number c in (0, 1) such that f(c) = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation ex = 3 − 2x, in the interval (0, 1)

Step-by-step explanation:

From the question we are told that

   The equation is f(x) = e^x - 3-2x  

    The interval is  [0, 1]

Generally f(0) is  

      f(0) = e^0 - 3-2(0)

=>   f(0) = 1 - 3-2(0)  

=>    f(0) = -2  

Generally f(1) is

       f(1) = e^1 - 3-2(1)

       f(1) = e^1 - 1

       f(1) = 1.718

From the value we see that at x =  0  , f(0) =  -2 which is below the x-axis

    and  the  at  x = 1  , f(1) =  1.718  which is above the x-axis

Now  the according to Intermediate Value Theorem , given the condition stated above, there will exist a root c in the interval  such that

    f(c) =  0  

3 0
3 years ago
4
yaroslaw [1]

Answer:

The expression that represents the given sequence is 5+6(n-1). Option C (not labeled).

Explanation:

<u>Arithmetic Sequences</u>

In an arithmetic sequence, each term can be obtained by adding or subtracting a fixed number to the previous term. That fixed number is called the common difference.

We are given the following sequence:

5, 11, 17, 23, 29, ...

Each term is located in a position starting from n=1. Let's test each option:

A For n=1 we should have the first term (5). Substituting n=1 into the general equation: 5+6(n+1) = 5+6(1+1) = 5+12 = 17. Since the resulting term is not 5, this option is incorrect.

B For n=1, 6+5(n+1)= 6+5(2)=16. This option is incorrect.

C (not labeled) For n=1, 5+6(n-1)=5+6(1-1)=5+0=5. The first term is correct. Let's test for the second term (n=2):

5+6(2-1)=5+6=11. Correct. For n=3

5+6(3-1)=5+12=17. Correct.

We can see the terms are increasing by 6, and the given sequence is also increasing by 6. Thus, This option is correct.

D For n=1, 6+5 (n-1)=6+0=6. This option is incorrect.

5 0
3 years ago
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