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lapo4ka [179]
3 years ago
10

During 30 minutes on the treadmill, LaToya burned 20 calories per minute. What is the rate of change of the function that repres

ents this situation?
10 calories per minute
20 calories per minute
30 calories per minute
50 calories per minute
Mathematics
1 answer:
Genrish500 [490]3 years ago
3 0
The answer I believe is 10 calories per minute because 20 x30 =600 I did that because she burned 20 calories per minute so with that being said there’s 60 minutes is the maximum minutes any time can go to so you do 600 divided by 60 which equals 10 which means she burns 10 calories per minute hope this is right
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The diagram shows a 5 cm x 5 cm x 5 cm cube.
mylen [45]

Answer:

~8.66cm

Step-by-step explanation:

The length of a diagonal of a rectangular of sides a and b is

\sqrt{a^2+b^2}

in a cube, we can start by computing the diagonal of a rectangular side/wall containing A and then the diagonal of the rectangle formed by that diagonal and the edge leading to A. If the cube has sides a, b and c, we infer that the length is:

\sqrt{\sqrt{a^2+b^2}^2 + c^2} = \sqrt{a^2+b^2+c^2}

Using this reasoning, we can prove that in a n-dimensional space, the length of the longest diagonal of a hypercube of edge lengths a_1, a_2, a_3, \ldots, a_n is

\sqrt{a_1^2 + a_2^2 + a_3^2 + \ldots + a_n^2}

So the solution here is

\sqrt{(5cm)^2 + (5cm)^2 + (5cm)^2} = \sqrt{75cm^2} = 5\sqrt{3cm^2} \approx 5\cdot 1.732cm = 8.66cm

5 0
3 years ago
Read 2 more answers
James decides to walk from school to the park to play ball. The park is only 0.5 mile from the school. He leaves school at 3:15
miskamm [114]

Answer: C

Step-by-step explanation: We see that it took James 10 minutes to go 0.5 miles, so to get how many miles an hour James is traveling, we need to get the miles to 1. To do this, multiply by 2, and this will give you that it took James 20 minutes to go 1 mile. This is option C.

7 0
3 years ago
The diagram below shows scalene triangle JKL. Triangle J K L. Side L K extends to point H to form exterior angle J K H. Which is
DiKsa [7]

Answer:

Step-by-step explanation:

  • Measure of angle J K H + measure of angle J K L = 180 degrees

True.

  • Measure of angle J K L + measure of angle J L K + measure of angle K J L = 180 degrees

True.  The sum of all interior angles = 180

  • Measure of angle J K H = measure of angle J L K =

Can't tell from info given.  

  • Measure of angle J K L + measure of angle J L K = 90 degrees

Can't tell from info given.

  • Measure of angle K J L = measure of angle K L J

Can't tell from info given.

8 0
3 years ago
Read 2 more answers
Determine if the equation below is a function with independent variable x. If so, find the domain. If not, find a value of x to
Andre45 [30]

Answer:

No because it contains points (3,4) and (3,-4).  You cannot have a x assigned to more than one y-value if you want it to be a function.

Step-by-step explanation:

A function has one output per input.

If we are trying to determine if the given is a function of x, then x is the input.

However I can get two outputs from plugging in x=3.

3^2+y^2=25

9+y^2=25

Subtract 9 on both sides:

y^2=25-9

y^2=16

Take the square root of both sides:

y=\pm \sqrt{16}

y=\pm 4.

So input x=3 yields y=4 and y=-4.

Since this input has more than one output then the given is not a function of x.

----Also!

If you graph the equation, it is a circle with radius 5 and center (0,0). So I could I plug in any number for x between -5 and 5 excluding -5 and 5 which would yield only one output each. Since plugging in either one gives:

(\pm 5)^2+y^2=25

25+y^2=25

Subtract 25 on both sides:

y^2=25-25

Simplify:

y^2=0

There is only one value y such that when you square it gives you 0. That is 0.

x=5 only gives y=0 and x=-5 only gives y=0.

There is no circle, unless it is a circle with radius 0 which means it really wouldn't be a circle, that is a function.

8 0
3 years ago
Solve. Find all solutions in [0,2).5 secx cotx+5 secx+ cotx+1=0
Dmitrij [34]
\begin{gathered} 5sec(x)cot(x)+5sec(x)+cot(x)+1=0 \\ Factoring \\ (cot(x)+1)(5sec(x)+1)=0 \\ cot(x)+1=0 \\ cot(x)=-1 \\ x=cot^{-1}(1) \\ x=\frac{3\pi}{4} \\  \\ 5sec(x)+1=0 \\ 5sec(x)=-1 \\ sec(x)=\frac{-1}{5},\text{ there is no solution} \\ Hence \\ All\text{ solution are }\frac{3\pi}{4}+\pi n,\text{ where n=1,2,3...} \\  \\  \end{gathered}

5 0
1 year ago
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