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2 years ago

What is ljnnjnjljlnlnjnjsxnjxnklxsnklnjdcjbcbncbjksacbkj 3/5 ÷ 2 1/2

Mathematics

1 answer:

NemiM [27]2 years ago

7 0

Answer:

6/25

Step-by-step explanation:

I think you're asking what 3/5 ÷ 2 1/2 is, but the gibberish throws me a bit for a loop.

Now to the problem:

$2 \frac{1}{2} = \frac{5}{2}$ so we can solve.

3/5 ÷ 5/2 =

$\frac{3}{5}$ · $\frac{2}{5}$ = $\frac{6}{25}$ (because division is equivalent to multiplacation by reciprocal)

So 6/25 is our answer.

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2 years ago

If Nina is N years old now and Maryna is M years old now, write the sum of their ages in 6 years

zimovet [89]

Answer:

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2 years ago

What expression is equivalent to 13-(-21)

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3 years ago

Read 2 more answers

Forces of 9 lbs and 13 lbs act at a 38º angle to each other. Find the magnitude of the resultant force and the angle that the re

fiasKO [112]

Answer: $R=20.84\ lb\quad 22.57^{\circ},15.43^{\circ}$

Step-by-step explanation:

Given

Two forces of 9 and 13 lbs acts $38^{\circ}$ angle to each other

The resultant of the two forces is given by

$\Rightarrow R=\sqrt{a^2+b^2+2ab\cos \theta}$

Insert the values

$\Rightarrow R=\sqrt{9^2+13^2+2(9)(13)\cos 38^{\circ}}\\\Rightarrow R=\sqrt{81+169+184.394}\\\Rightarrow R=\sqrt{434.394}\\\Rightarrow R=20.84\ lb$

Resultant makes an angle of

$\Rightarrow \alpha=\tan^{-1}\left( \dfrac{b\sin \theta}{a+b\cos \theta}\right)\\\\\text{Considering 9 lb force along the x-axis}\\\\\Rightarrow \alpha =\tan^{-1}\left( \dfrac{13\sin 38^{\circ}}{9+13\cos 38^{\circ}}\right)\\\\\Rightarrow \alpha =\tan^{-1}(\dfrac{8}{19.244})\\\\\Rightarrow \alpha=22.57^{\circ}$