Larry joggers 2 3/5 km. How many kilometers and meters did he jog?

Mathematics

1 answer:

Montano1993 [528]4 years ago

3 0

Answer: The answer is 2 kilometres and 600 metres.

Step-by-step explanation: The distance jog by Larry is given by

$d=2\dfrac{3}{5}~\textup{kilometres}.$

We are to find the distance jog by Larry in kilometres and metres.

We know that

1 kilometres = 1000 metres.

The conversion is as follows:

$d\\\\=2\dfrac{3}{5}~\textup{kilometres}\\\\=2~\textup{kilometres}\dfrac{3}{5}\times 1000~\textup{metres}\\\\=2~\textup{kilometres}~600~\textup{metres}.$

Thus, the answer is 2 kilometres and 600 metres.

You might be interested in

-0.5 0 1 -2 1.5 Where is Point D on the number line? ?

Over [174]

Answer:

Its -1.9 for where point D is

4 0

3 years ago

Read 2 more answers

Raphael paid $167 for a camera during a 75% off sale. what was the cameras regular price?

mihalych1998 [28]

As the camera was at 25% of the original price, this means you've got to multiply this number to by 4 to get the original price.
167*4= 668
Therefore, the camera's regular price is $668
Hope this helps :)

6 0

3 years ago

Read 2 more answers

Can someone help me ???

Sphinxa [80]

Rational numbers are whole integers w no fractions. bc the square root of 72 is not a whole number, it is irrational

4 0

3 years ago

How do i prove 2sec^2x-2sec^2xsin^2-sin^2x-cos^2x=1

kramer

$2\sec^2x-2\sec^2x\sin^2x-\sin^2x-\cos^2x=2\sec^2x(1-\sin^2x)-(\sin^2x+\cos^2x)$

Since $\sin^2x+\cos^2x=1$ , you have

$2\sec^2x(1-\sin^2x)-(\sin^2x+\cos^2x)=2\sec^2x\cos^2x-1$

and since $\sec x=\dfrac1{\cos x}$ , you end up with $2-1=1$ .