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according to the line plot what is the total distance that was run by Runners who each ran for 1/3 of a mile​
according to the line plot what is the total distance - 1

Answers

Answer 1
Answer:

The total distance that was run by Runners who each ran for 1/3 of a mile​ will be 4/3.

What is distance?

The distance is defined as the length of the space between the two points separated from each other.

From the graph, the distance of 1/3 miles is covered by 4 runners so the total distance will be calculated as:-

Distance = 1/3  x  4

Distance = 4/3  miles

Hence the total distance that was run by Runners who each ran for 1/3 of a mile​ will be 4/3.

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Answer 2
Answer:

So you would do 1/3 times 4 so that is 1 and 1/3

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Answer:

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A piece of rope there is 28 feet long is cut into two pieces. One is use to form a circle and others used to form a square. Write a function G representing the area of the square as a function of the radius of the circle R

Answers

The function g representing the area of the square as a function of the radius of the circle r is given as:

g(r) = 49 - 22r + (121r^2)/(49)

Solution:

Given that,

length of rope = 28 feet

Let "c" be the circumference of circle

Let "p" be the perimeter of square

Therefore,

length of rope = circumference of circle + perimeter of square

c + p = 28 ------- eqn 1

The circumference of circle is given as:

c = 2 \pi r

Where, "r" is the radius of circle

Substitute the above circumference in eqn 1

2 \pi r + p = 28

p = 28 - 2 \pi r ----------- eqn 2

If "a" is the length of each side of square, then the perimeter of sqaure is given as:

p = 4a

Substitute p = 4a in eqn 2

4a = 28 - 2 \pi r\n\na = (28 - 2 \pi r)/(4)\n\na = 7 - ( \pi r)/(2)

The area of square is given as:

area = (side)^2\n\narea = a^2

Substitute the value of "a" in above area expression

area = (7 - ( \pi r)/(2))^2  ------ eqn 3

We know that,

(a - b)^2 = a^2 - 2ab + b^2

Therefore eqn 3 becomes,

area = 7^2 -2(7)((\pi r)/(2)) + (( \pi r)/(2))^2\n\narea = 49 - 7 \pi r + ( (\pi)^2 r^2 )/(4)

\text{ substitute } \pi = (22)/(7)

area = 49 - 7 * (22)/(7) * r + ((22)/(7))^2 * (r^2 )/(4)\n\narea = 49 - 22r + (121)/(49) * r^2\n\narea = 49 - 22r + (121r^2)/(49)

Let g(r) represent the area of the square as a function of the radius of the circle r, then we get

g(r) = 49 - 22r + (121r^2)/(49)

Thus the function is found

Hurry I am running out of time! I have 10 minutes.

Answers

What it doooooooooooooo

Is {(98,6),(99,0),(100,7)} a function?

Answers

Answer:

Yes , this is the function

Step-by-step explanation:

Because every domain has their only one image

-5+i/2i how do I break this down?​

Answers

Answer:

i2 = -1

Step-by-step explanation:

5i ⋅i⋅(−2i)= −

10 ⋅ i2 ⋅ i= − 10 ⋅ (−1) ⋅ i = 10i

Solve the matrix equation for a, b, c, and d. [1 2] [a b] [6 5][3 4] [c d]= [19 8]

Answers

Answer:

The answer is "\bold{\left[\begin{array}{cc}a&b\nc&d\end{array}\right] = \left[\begin{array}{cc}7&-2\n -(1)/(2)&(7)/(2)\end{array}\right]}".

Step-by-step explanation:

\bold{\left[\begin{array}{cc}1&2\n3&4\end{array}\right] \left[\begin{array}{cc}a&b\nc&d\end{array}\right] = \left[\begin{array}{cc}6&5\n 19&8\end{array}\right]}

Solve the L.H.S part:

\left[\begin{array}{cc}1&2\n3&4\end{array}\right] \left[\begin{array}{cc}a&b\nc&d\end{array}\right]\n\n\n\left[\begin{array}{cc}a+2c&b+2d\n3a+4c&3b+4d\end{array}\right]

After calculating the L.H.S part compare the value with R.H.S:

\left[\begin{array}{cc}a+2c&b+2d\n3a+4c&3b+4d\end{array}\right]= \left[\begin{array}{cc}6&5\n 19&8\end{array}\right]} \n\n

\to a+2c =6....(i)\n\n\to b+2d =5....(ii)\n\n\to 3a+4c =19....(iii)\n\n\to 3b+4d = 8 ....(iv)\n\n

In equation (i) multiply by 3 and subtract by equation (iii):

\to 3a+6c=18\n\to 3a+4c=19\n\n\text{subtract}... \n\n\to 2c = -1\n\n\to c= - (1)/(2)

put the value of c in equation (i):

\to a+ 2 (- (1)/(2))=6\n\n\to a- 2 * (1)/(2)=6\n\n\to a- 1=6\n\n\to a =6 +1\n\n\to a = 7\n

In equation (ii) multiply by 3 then subtract by equation (iv):

\to 3b+6d=15\n\to 3b+4d=8\n\n\text{subtract...}\n\n\to 2d = 7\n\n\to d= (7)/(2)\n

put the value of d in equation (iv):

\to 3b+4 ((7)/(2))=8\n\n\to 3b+4 * (7)/(2)=8\n\n\to 3b+14=8\n\n\to 3b =8-14\n\n\to 3b = -6\n\n\to b= (-6)/(3)\n\n\to b= -2

The final answer is "\bold{\left[\begin{array}{cc}a&b\nc&d\end{array}\right] = \left[\begin{array}{cc}7&-2\n -(1)/(2)&(7)/(2)\end{array}\right]}".
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