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3 years ago

The graph represents a person’s heart rate in beats per minute during 30 minutes of exercise.

Mathematics

2 answers:

Ronch [10]3 years ago

8 0

Answer:

The Answer is C: The heart rate increases for 6 minutes, remains constant for 19 minutes, and then gradually decreases for 5 minutes. :)

Juliette [100K]3 years ago

4 0

GOOD

LUCK

CHUCK

:))))))))

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19 minus 14 5/8hjkvcbkifc

mars1129 [50]

If we don't count the letters, $19 - 14\frac{5}{8}$ The answer would be: 4 3/8! Hope this helped! Brainliest is always appreciated!

7 0

2 years ago

3. Find the slope between the two points (-3,3) and (3,-1) Your Answer:

WARRIOR [948]

The slope is -2/3 I hope this helped

4 0

2 years ago

Read 2 more answers

60 EXTRA POINTS!!!! I need help with writing an inequality for a graph. It is below. :)

ch4aika [34]

Answer:

The answer to your question is: y ≥ - x + 6 and y ≥ x - 4

Step-by-step explanation:

Look of a pairs of points for each graph

Line 1 Line 2

A(0, 6) B (5, 1) C (7,3) D(5, 1)

m = (1- 6) / (5- 0) m = (1 - 3) / (5 - 7)

m = -5/5 = -1 m = -2 / -2 = 1

Find the line equation for each slope (y - y1) = m (x - x1)

(y - 1) = -1 (x - 5) (y - 1) = 1 (x - 5)

y - 1 = -x + 5 y -1 = x - 5

y = -x + 5 + 1 y = x - 5 + 1

y = -x + 6 y = x - 4

Inequality

y ≥ - x + 6 y ≥ x - 4

8 0

2 years ago

You have D dollars to buy fence to enclose a rectangular plot of land (see figure at right). The fence for the top and bottom co

alex41 [277]

The perimeter of the rectangular plot of land is given by the expression below

$P=2x+2y$

On the other hand, since the available money to buy fence is D dollars,

$\begin{gathered} D=4(2x)+3(2y) \\ \Rightarrow D=8x+6y \\ D\rightarrow\text{ constant} \end{gathered}$

Furthermore, the area of the enclosed land is given by

$A=xy$

Solving the second equation for x,

$\begin{gathered} D=8x+6y \\ \Rightarrow x=\frac{D-6y}{8} \end{gathered}$

Substituting into the equation for the area,

$\begin{gathered} A=(\frac{D-6y}{8})y \\ \Rightarrow A=\frac{D}{8}y-\frac{3}{4}y^2 \end{gathered}$

To find the maximum possible area, solve A'(y)=0, as shown below

$\begin{gathered} A^{\prime}(y)=0 \\ \Rightarrow\frac{D}{8}-\frac{3}{2}y=0 \\ \Rightarrow\frac{3}{2}y=\frac{D}{8} \\ \Rightarrow y=\frac{D}{12} \end{gathered}$

Therefore, the corresponding value of x is

$\begin{gathered} y=\frac{D}{12} \\ \Rightarrow x=\frac{D-6(\frac{D}{12})}{8}=\frac{D-\frac{D}{2}}{8}=\frac{D}{16} \end{gathered}$ <h2>Thus, the dimensions of the fence that maximize the area are x=D/16 and y=D/12.</h2><h2>As for the used money,</h2> $\begin{gathered} top,bottom:\frac{8D}{16}=\frac{D}{2} \\ Sides:\frac{6D}{12}=\frac{D}{2} \end{gathered}$ <h2>Half the money was used for the top and the bottom, while the other half was used for the sides.</h2>

7 0

8 months ago

Which of the statements is also true in Spherical geometry?

Rus_ich [418]

I think all of them are true

3 0

3 years ago