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Sonja [21]
3 years ago
15

You exercise for one hour each day and you burn 8.5 calories per minute. If you graphed the number of calories you burned as the

output, compared to the number of minutes you exercise as input, what would the graph look like? Which of the following would not be true?A.The graph would be a line.B.The graph would rise from left to right.C.The graph would be only in the first quadrant.D.The larger the number of minutes, the fewer calories you would burn.
Mathematics
1 answer:
mezya [45]3 years ago
4 0
For every minute you exercise you burn 8.5 calories. It's a linear pattern, so the graph would be a line.
The more minutes you exercise, the more calories you burn, so the graph would rise from left to right.
The number of minutes and the number of calories can't be negative numbers, so the graph would be only in the first quadrant.
The larger the number of minutes, the more calories you would burn.

The answer is D.
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hammer [34]

the answer is B

because 54x 3= 162 x3 = 486x3= 1458x3=4,374x3=13,122x3=39,366x3=118,098x3=354,294

7 0
3 years ago
15% of Alina’s workout consisted of cardio exercise. 25% of Dane’s workout consisted of cardio exercise. Which statement must be
Arisa [49]


The correct answer is B. If Alina’s workout was 80 minutes, and Dane’s workout was 48 minutes, they spent the same amount of time on cardio exercise.

To figure this out, first find 15% of 80 and then 25% of 48.

To find 15% of 80, you can first find 10% of 80 which is 8(just move the decimal one place to the left). The remaining 5% will be half of 8 since 8 is 10%. Add both the numbers up and you will get 12. (8+4=12)

To find 25% of 48, you can divide 48 by 4. Since percent means 'out of 100', 25% is 25 of 100. 100 divided by 25 is equal to 4. So, 48 divided by 4 is equal to 12.

So, the final answer is B. They both equal to 12.

Hope it helps :)

6 0
3 years ago
A box is to be constructed from a sheet of cardboard that is 10 cm by 60 cm by cutting out squares of length x by x from each co
Stells [14]

Volume of a rectangular box = length x width x height<span>

From the problem statement,
length = 60 - 2x
width = 10 - 2x
height = x</span>

<span>
where x is the height of the box or the side of the equal squares from each corner and turning up the sides

V = (60-2x) (10-2x) (x)
V = (60 - 2x) (10x - 2x^2)
V = 600x - 120x^2 -20x^2 + 4x^3
V = 4x^3 - 100x^2 + 600x

To maximize the volume, we differentiate the expression of the volume and equate it to zero.

V = </span>4x^3 - 100x^2 + 600x<span>
dV/dx = 12x^2 - 200x + 600
12x^2 - 200x + 600 = 0</span>

<span>x^2 - 50/3x + 50 = 0

Solving for x,
x1 = 12.74 ;      Volume = -315.56  (cannot be negative)
x2 = 3.92 ;      Volume = 1056.31

So, the answer would be that the maximum volume would be 1056.31 cm^3.
</span><span>
</span>
6 0
3 years ago
Find the solution of the given initial value problem:<br><br> y''- y = 0, y(0) = 2, y'(0) = -1/2
igor_vitrenko [27]

Answer:  The required solution of the given IVP is

y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.

Step-by-step explanation:  We are given to find the solution of the following initial value problem :

y^{\prime\prime}-y=0,~~~y(0)=2,~~y^\prime(0)=-\dfrac{1}{2}.

Let y=e^{mx} be an auxiliary solution of the given differential equation.

Then, we have

y^\prime=me^{mx},~~~~~y^{\prime\prime}=m^2e^{mx}.

Substituting these values in the given differential equation, we have

m^2e^{mx}-e^{mx}=0\\\\\Rightarrow (m^2-1)e^{mx}=0\\\\\Rightarrow m^2-1=0~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mx}\neq0]\\\\\Rightarrow m^2=1\\\\\Rightarrow m=\pm1.

So, the general solution of the given equation is

y(x)=Ae^x+Be^{-x}, where A and B are constants.

This gives, after differentiating with respect to x that

y^\prime(x)=Ae^x-Be^{-x}.

The given conditions implies that

y(0)=2\\\\\Rightarrow A+B=2~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)

and

y^\prime(0)=-\dfrac{1}{2}\\\\\\\Rightarrow A-B=-\dfrac{1}{2}~~~~~~~~~~~~~~~~~~~~~~~~(ii)

Adding equations (i) and (ii), we get

2A=2-\dfrac{1}{2}\\\\\\\Rightarrow 2A=\dfrac{3}{2}\\\\\\\Rightarrow A=\dfrac{3}{4}.

From equation (i), we get

\dfrac{3}{4}+B=2\\\\\\\Rightarrow B=2-\dfrac{3}{4}\\\\\\\Rightarrow B=\dfrac{5}{4}.

Substituting the values of A and B in the general solution, we get

y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.

Thus, the required solution of the given IVP is

y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.

4 0
3 years ago
chris can run at a constant speed of 12 km/h. How ling will it take him to run from his home to the park, a distance of 0.8 km?
densk [106]

Recall that distance = speed times time. Thus, time = distance / speed.

Here:

time = distance / speed = 0.8 km / 12 km/hr = 1/15 hr.

Note that 1/15 hr = 1/15 (60 min) = 4 minutes

8 0
3 years ago
Read 2 more answers
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