All categories

3 years ago

Anyone know if -2.35/30.8 is equal to -47/616

Mathematics

1 answer:

Dahasolnce [82]3 years ago

3 0

It is both equations are equal to -0.076.

You might be interested in

If lisa can run a mile in 7 minutes. At that rate of speed, how long would it take her to run 2 kilometers?

murzikaleks [220]

Answer:

8.75

Step-by-step explanation:

1 mile = 1.6 km

So, every 7 minutes, 1.6 km is ran.

Divide 1.6 by 2 to figure out how much more she is running.

1.6 / 2 = 0.8

Divide the known pace by the extra distance to figure out the time

7 min / 0.8 km = 8.75 min

5 0

3 years ago

Lock #1 Answer the question below to UNLOCK the next room. (TYPE THE WORD: ALL

Ira Lisetskai [31]

Answer:

UNLOCK

A normal number

8 0

3 years ago

Read 2 more answers

When ∆A′B′C′ is reflected across the line x = -2 to form ∆A″B″C″, vertex (note: please enter your response using apostrophes ins

maria [59]

(A) because it is reflecting over the x-axis.

7 0

3 years ago

Help Friends, Ayuda Amigos :(.

mojhsa [17]

Answer:

i dont speak spanish

Step-by-step explanation:

sorry i dont understand you...

4 0

3 years ago

D/d{cosec^-1(1+x²/2x)} is equal to

SIZIF [17.4K]

Step-by-step explanation:

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:\dfrac{d}{dx} {cosec}^{ - 1} \bigg( \dfrac{1 + {x}^{2} }{2x} \bigg)$

Let assume that

$\rm :\longmapsto\:y = {cosec}^{ - 1} \bigg( \dfrac{1 + {x}^{2} }{2x} \bigg)$

We know,

$\boxed{\tt{ {cosec}^{ - 1}x = {sin}^{ - 1}\bigg( \dfrac{1}{x} \bigg)}}$

So, using this, we get

$\rm :\longmapsto\:y = sin^{ - 1} \bigg( \dfrac{2x}{1 + {x}^{2} } \bigg)$

Now, we use Method of Substitution, So we substitute

$\red{\rm :\longmapsto\:x = tanz \: \rm\implies \:z = {tan}^{ - 1}x}$

So, above expression can be rewritten as

$\rm :\longmapsto\:y = sin^{ - 1} \bigg( \dfrac{2tanz}{1 + {tan}^{2} z} \bigg)$

$\rm :\longmapsto\:y = sin^{ - 1} \bigg( sin2z \bigg)$

$\rm\implies \:y = 2z$

$\bf\implies \:y = 2 {tan}^{ - 1}x$

So,

$\bf\implies \: {cosec}^{ - 1}\bigg( \dfrac{1 + {x}^{2} }{2x} \bigg) = 2 {tan}^{ - 1}x$

Thus,

$\rm :\longmapsto\:\dfrac{d}{dx} {cosec}^{ - 1} \bigg( \dfrac{1 + {x}^{2} }{2x} \bigg)$

$\rm \: = \: \dfrac{d}{dx}(2 {tan}^{ - 1}x)$

$\rm \: = \: 2 \: \dfrac{d}{dx}( {tan}^{ - 1}x)$

$\rm \: = \: 2 \times \dfrac{1}{1 + {x}^{2} }$

$\rm \: = \: \dfrac{2}{1 + {x}^{2} }$

Hence, 

$\purple{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx} {cosec}^{ - 1} \bigg( \dfrac{1 + {x}^{2} }{2x} \bigg) = \frac{2}{1 + {x}^{2} }}}}$

Hence, Option (d) is correct.

6 0

2 years ago