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3 years ago

12

Ram,raju and ravi can finish the running competition on circular track in 15 mins,20mins and 30mins respectively .If they start

together at 7.00am from the starting point,at what time they will meet together again at th starting point

1 answer:

OlgaM077 [116]3 years ago

5 0

About 8:00 is when they will all have met up ram will have ran 4 laps raju will have ran 3 laps and ravi will have ran 2 laps

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Triangle JKL has an area of in2 and a perimeter of in.

S_A_V [24]

Answer:

Area = 24 in²

Perimeter = 24in

Step-by-step explanation:

To find the area given the length of the 3 sides:

a= 8in

b= 6.4in + 3.6in= 10in

c= 6in

Add them together:

8in + 10in + 6in

= 24in²

To find the perimeter you do the same thing:

a= 8in

b= 10in

c= 6in

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What is Derricka's goal, in miles, for the total distance she will swim on Monday, Tuesday, and Wednesday? Show your work.

denpristay [2]

Answer:

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Step-by-step explanation:

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3 years ago

The answer and explanation to question 1

xxMikexx [17]

Answer:

20, 31, 33, 35, 55, 57, 58, 59, 72, 73, 79, 86, 87, 88

Step-by-step explanation:

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20, 31, 33, 35, 55, 57, 58, 59, 72, 73, 79, 86, 87, 88

7 0

3 years ago

The area of a right triangle is sixty square centimetres. Its base is one centimetre less than twice its height. If the base and

NemiM [27]

Answer:

5. None of the above

Step-by-step explanation:

We have this information:

A = 60 cm2

b = 2*h - 1

Let's find out b and h:

The equation of the are of a triangle is A = (b*h)/2, if we replace it with our information, we have

$60 = ((2*h - 1) * h)/2\\ 120 = 2*h^2 - h\\ 0 = 2*h^2 - h - 120$

Let's find h with the quadratic formula:

$\fbox {Quadratic formula:} \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$h = \frac{-(-1)\pm\sqrt{(-1)^2-4*2*(-120)}}{2*2} = \frac{1\pm\sqrt{961}}{4} = \frac{1\pm\ 31}{4}$

h = 8 or h = -7.5

But h represents the height of the triangle, so it has to be a positive number, that's h = 8.

If we replace this in the equation we had for b, we have that b = 2*8 - 1 = 15.

Now we can calculate the hypotenuse with the Pythagorean equation

$\fbox {Pythagorean equation:}$ <em>The square of the length of the hypotenuse (the side opposite the right angle) of a right triangle is equal to the sum of the squares of the two legs (the two sides that meet at a right angle). </em>

The base and the height are our legs. We will use "H" for the hypotenuse

$H^2 = b^2 + h^2 = 15^2 + 8^2 = 289\\ H = \sqrt {289} = 17$

H = 17

If we decrease the base and the height by 2 centimeters, we have

b' = 15 - 2 = 13 and h' = 8 - 2 = 6

With this, let's calculate the new hypotenuse:

$H'^2 = b'^2 + h'^2 = 13^2 + 6^2 = 205\\ H' = \sqrt {205} \approx 14.3$

$H' \approx 14.3$

So, the hypotenuse decreases $H - H' \approx 17 - 14.3 = 2.7$

3 0

3 years ago

Which of the following rational numbers is the largest? 0.9 –8/9 –2 1/2 9/4

puteri [66]

-8/9 I think sorry if I’m wrong

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3 years ago