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klemol [59]
3 years ago
13

4 people can trim a hedge in one hour.

Mathematics
2 answers:
Pani-rosa [81]3 years ago
6 0
4 people in 60 minutes =
1 person in 15 minutes

3 people = 45 minutes
faltersainse [42]3 years ago
6 0

Answer:

The total time taken by 3 men to trim the hedge would be 80 minutes.

Step-by-step explanation:

1) The time taken by 4 men to trim the hedge is 1 hour or 60 min.

2) One me would take 4 times more time alone i.e. Time taken by only one man would be 4 hours or 240 min.

3) The time taken by 3 men would be 3 times less than time taken by one man i.e. 240/3=80 min.

#of people = 4

time taken for 4 ppl = 1hr / 60mins  

# of people = 3

time taken = 1/x

4 - 1/60

3 - 1/x

cross multiply

x=60/3*4

x=80

hope this helps :)

mark me brainliest :D

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What is the value of x? There is isosceles triangle. The measure of angle which is between two congruent sides is (6x+10⁰). The
max2010maxim [7]
<h3><u>Answer</u><u>:</u><u>-</u></h3>

x=17

<h3><u>step-by</u><u>-</u><u>step</u><u> </u><u>Explanation</u><u>:</u><u>-</u></h3>

{\boxed{\quad\:\mapsto\rm Firstly\:Let's\:understand\:the\:concept:-}}

<h3 />

This is a isosceles triangle. As it is a triangle we can apply sum theory. we have to take the sum of given unknown polynomials as 180° .Then by solving it we can find the value of x.

<h3><u>Solution</u><u>:</u><u>-</u></h3>

Given angles

  • (6x+10°)
  • (x+17°)
  • (4x-34)°

According to sum theory

{\boxed{\sf The \:sum\:of\:angles=180°}}

  • Substitute the values

\qquad \quad{:}\longmapsto\tt (6x+10)+(x+17)+(4x-34)=180

  • Remove brackets

\qquad \quad{:}\longmapsto\tt 6x+10+x+17+4x-34=180

  • Together like polynomials and constants

\qquad \quad{:}\longmapsto\tt 6x+x+4x+10+17-34=180

\qquad \quad{:}\longmapsto\tt 11x-7=180

  • Interchange sides

\qquad \quad{:}\longmapsto\tt 11x=180+7

\qquad \quad{:}\longmapsto\tt 11x=187

\qquad \quad{:}\longmapsto\tt x=\cancel{\dfrac{187}{11}}

  • Simplify

\qquad \quad{:}\longmapsto\tt x=17

\therefore\sf The \:value\:of\:x\;is\:17.

8 0
3 years ago
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