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tatyana61 [14]
3 years ago
13

a motor scooter travels 20 miles an hour in the same time that a bicycle covers 9 miles an hour if the rate of the scooter is 9

miles per hour more than twice the rate of the bicycle find both rates?
Mathematics
1 answer:
sdas [7]3 years ago
3 0
0.4 hours or 24 minutes
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yanalaym [24]

Answer:

Negative 2.

Step-by-step explanation:

2x -1 would have an outcome of a negative number since 2 is positive and one is negative. if they were both positive numbers or negative numbers then the answer would be positive

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The length of a rectangle is 5 metres less than twice the breadth. If the perimeter is 50 meters,find the length and breadth
MAXImum [283]
<h3><u>S</u><u> </u><u>O</u><u> </u><u>L</u><u> </u><u>U</u><u> </u><u>T</u><u> </u><u>I</u><u> </u><u>O</u><u> </u><u>N</u><u> </u><u>:</u></h3>

As per the given question, it is stated that the length of a rectangle is 5 m less than twice the breadth.

Assumption : Let us assume the length as "l" and width as "b". So,

\twoheadrightarrow \quad\sf{ Length =2(Width)-5}

\twoheadrightarrow \quad\sf{ \ell=(2b-5) \; m}

Also, we are given that the perimeter of the rectangle is 50 m. Basically, we need to apply here the formula of perimeter of rectangle which will act as a linear equation here.

\\ \twoheadrightarrow \quad\sf{ Perimeter_{(Rectangle)} = 2(\ell +b) } \\

  • <em>l</em> denotes length
  • <em>b</em> denotes breadth

\\ \twoheadrightarrow \quad\sf{50= 2(2b-5+b)} \\

\\ \twoheadrightarrow \quad\sf{50= 2(3b-5)} \\

\\ \twoheadrightarrow \quad\sf{50= 6b - 10} \\

\\ \twoheadrightarrow \quad\sf{50+10= 6b} \\

\\ \twoheadrightarrow \quad\sf{60= 6b} \\

\\ \twoheadrightarrow \quad\sf{\cancel{\dfrac{60}{6}}=b} \\

\\ \twoheadrightarrow \quad\underline{\bf{10\; m = Width }} \\

Now, finding the length. According to the question,

\twoheadrightarrow \quad\sf{ \ell=(2b-5) \; m}

\twoheadrightarrow \quad\sf{ \ell=2(10)-5\; m}

\twoheadrightarrow \quad\sf{ \ell=20-5\; m}

\\ \twoheadrightarrow \quad\underline{\bf{15\; m = Length }} \\

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3 years ago
Two pieces of lengths 13/3 m and 3/5 m are cut off from a rope 15
garri49 [273]

Hello there! ;) The way you solve this type of problem is you must find common denominators. So simplify 13/3= 4 1/3. So we can do 1/3*5/5=5/15. Next we must do 3/5*3/3= 9/15. So add 5/15+9/15= 14/15. And add the 4+ 14/15=          4 14/15. So then you must subtract 15 - 4 14/15= 10 1/15 I believe. I believe 10 1/15 is the remainder of the rope. You might want to check in with someone else though.

Hope I help! :D

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Need help with this problem:
arlik [135]

Answer:

843 days

Step-by-step explanation:

7 0
4 years ago
How do you solve this?
Fantom [35]

Am sorry but it think you have forgotten to write the equation or to add an image

6 0
3 years ago
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