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S_A_V
7 days ago
10

The school playground is in the shape of a pentagon. There is a drinking fountain at each of the 5 corners of the playground. Th

ere is a path that connects each water fountain. How many ways can someone walk from one drinking fountain to another drinking fountain without retracing their steps and while staying on the path?
Mathematics
1 answer:
PIT_PIT [9.1K]7 days ago
7 0
The answer is 21 distinct paths.
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Estimate the sum of 196 and 482
Svet_ta [9496]

we understand that

A fast method for estimating the total of two numbers involves rounding each number and adding the rounded results.

case a) We round to the nearest hundred

so

196 Round up equals 200

482 Round up equals 500

Calculate the estimated sum

200+500=700

case b) We round to the nearest ten

so

196 Round up equals 200

482 Round down equals 480

Calculate the estimated sum

200+480=680

8 0
18 days ago
Read 2 more answers
Terrence walks at a pace of 2 mi/h to the theater and watches a movie for 2 hours and 15 min he rides a bus back home at 40 mi/h
zzz [9073]

Solution:

Let x be the distance in miles from the house to the theater.

Total time taken for the trip =3.5-2.25=1.25 Hr

Then average speed =\frac{Distance}{Time}= \frac{2x}{1.25}\\

Terrence walks at a speed of 2 miles per hour to the theater and returns home at 40 miles per hour.

Thus, Average Speed =\frac{2+40}{2}= 21\\ mi/hr

Both average speeds must be equivalent. Hence, we can express this as:

\frac{2x}{1.25}=21\\
\\
2x=1.25*21\\
\\
x=13.125\\

Consequently, the distance from home is 13.125 miles.

4 0
27 days ago
Find the midpoint of the segment with endpoints of −11 + i and −4 + 4i.
tester [8820]

\bf \begin{cases} -11+i\implies -11+1i\\ \qquad (-11,1)\\ -4+4i\\ \qquad (-4,4) \end{cases} \\\\\\ ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-11}~,~\stackrel{y_1}{1})\qquad (\stackrel{x_2}{-4}~,~\stackrel{y_2}{4}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~,~~~ \cfrac{ y_2 + y_1}{2} \right)


\bf \left( \cfrac{-4-11}{2}~~,~~ \cfrac{4+1}{2}\right)\implies \left(-\frac{15}{2}~,~\frac{5}{2} \right)\implies \left(-7\frac{1}{2}~,~2\frac{1}{2} \right) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill -7\frac{1}{2}+2\frac{1}{2}i~\hfill

5 0
17 days ago
Read 2 more answers
How did the beetle uncover the ants secret plan?
PIT_PIT [9117]
It recorded its conversation.
8 0
11 days ago
Which of these equations have no solution? Check all that apply. 2(x + 2) + 2 = 2(x + 3) + 1 2x + 3(x + 5) = 5(x – 3) 4(x + 3) =
Inessa [8997]

Answer:

  • 2(x + 2) + 2 = 2(x + 3) + 1
  • 2x + 3(x + 5) = 5(x – 3)
  • 5(x + 4) – x = 4(x + 5) – 1

Step-by-step explanation:

To identify solutions, subtract the right side from both sides and simplify.

1. For 2(x + 2) + 2 = 2(x + 3) + 1

   Subtract right side: 2(x + 2) + 2 - (2(x + 3) + 1) = 0

   Simplify: 2x + 4 + 2 - 2x - 6 - 1 = 0

   Which results in -1 = 0, meaning no solutions.

__

2. For 2x + 3(x + 5) = 5(x – 3)

   Subtract right side: 2x + 3(x + 5) - 5(x – 3) = 0

   Simplify: 2x + 3x + 15 - 5x + 15 = 0

   This leads to 30 = 0, no solutions again.

__

3. Equation 4(x + 3) = x + 12

   Subtract right side: 4(x + 3) - (x + 12) = 0

   Simplify: 4x + 12 - x - 12 = 0

   Equals 3x = 0, so one solution: x = 0.

__

4. Equation 4 – (2x + 5) = (–4x – 2)

   Subtract right side: 4 – (2x + 5) - (–4x – 2) = 0

   Simplify: 4 - 2x - 5 + 4x + 2 = 0

   Gives 2x + 1 = 0, meaning one solution: x = -1/2.

__

5. Equation 5(x + 4) – x = 4(x + 5) – 1

   Subtract right side: 5(x + 4) – x - (4(x + 5) – 1) = 0

   Simplify: 5x + 20 - x - 4x - 20 + 1 = 0

   Results in 1 = 0, no solution here.

3 0
1 month ago
Read 2 more answers
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