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3 years ago

How many 3 letter words can be made from 4 letters “FGHI” if repetition is allowed and if repetition isn’t alllowed ?

Mathematics

1 answer:

PolarNik [594]3 years ago

8 0

Answer:

1. Word: Fig

Definition: The hollow, pear-shaped false fruit of the fig tree, with sweet, pulpy flesh containing numerous tiny, seed-like true fruits.

2. Word: GIF

Definition: Is an image file that is compressed to allow it to be transferred quickly, or an animated gif which is a collection of images played in sequence to appear to move.

3. Word: Ghi

Definition: Uncountable (is also the alternative spelling of ghee)

Step-by-step explanation:

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Update:

ohaa [14]

Answer:

Love vines!

Step-by-step explanation:

But vine left us while getting milk. And tick tock is like our step father that we've grew to love :)

4 0

3 years ago

Read 2 more answers

What is the product? Two-thirds times eight-ninths StartFraction 10 over 27 EndFraction StartFraction 16 over 27 EndFraction Two

melomori [17]

Answer:

16/27

Step-by-step explanation:

First let's convert this info an equation and then we will solve.

(2/3) * (8/9) = ?

( 2 * 8 ) / ( 3 * 9 ) = ?

( 16 ) / ( 27 ) = ?

16 / 27 = ?

So our fractional answer is 16/27.

In words this would be sixteen over twenty seven.

Cheers.

3 0

2 years ago

2 * 2 * 4 * 5 * 6 * 2 * 2 * 5 * 7 * 5 = ? will mark brainliest!

BaLLatris [955]

Answer:

336,000

Step-by-step explanation:

2 × 2 = 4 x 4 = 16 × 5 = 80 × 6 = 480 x 2 = 960 x 2 = 1920 × 5 = 9600 x 7 = 62700 × 5 = 336,000

3 0

2 years ago

8 to 14 as a ratio as a fraction in simplest form.

padilas [110]

Answer:

4/7 or 4:7

Step-by-step explanation:

8/14 divided by 2= 4/7

3 0

3 years ago

Read 2 more answers

Solve only if you know the solution and show work.

SashulF [63]

$\displaystyle\int\frac{\cos x+3\sin x+7}{\cos x+\sin x+1}\,\mathrm dx=\int\mathrm dx+2\int\frac{\sin x+3}{\cos x+\sin x+1}\,\mathrm dx$

For the remaining integral, let $t=\tan\dfrac x2$ . Then

$\sin x=\sin\left(2\times\dfrac x2\right)=2\sin\dfrac x2\cos\dfrac x2=\dfrac{2t}{1+t^2}$
$\cos x=\cos\left(2\times\dfrac x2\right)=\cos^2\dfrac x2-\sin^2\dfrac x2=\dfrac{1-t^2}{1+t^2}$

and

$\mathrm dt=\dfrac12\sec^2\dfrac x2\,\mathrm dx\implies \mathrm dx=2\cos^2\dfrac x2\,\mathrm dt=\dfrac2{1+t^2}\,\mathrm dt$

Now the integral is

$\displaystyle\int\mathrm dx+2\int\frac{\dfrac{2t}{1+t^2}+3}{\dfrac{1-t^2}{1+t^2}+\dfrac{2t}{1+t^2}+1}\times\frac2{1+t^2}\,\mathrm dt$

The first integral is trivial, so we'll focus on the latter one. You have

$\displaystyle2\int\frac{2t+3(1+t^2)}{(1-t^2+2t+1+t^2)(1+t^2)}\,\mathrm dt=2\int\frac{3t^2+2t+3}{(1+t)(1+t^2)}\,\mathrm dt$

Decompose the integrand into partial fractions:

$\dfrac{3t^2+2t+3}{(1+t)(1+t^2)}=\dfrac2{1+t}+\dfrac{1+t}{1+t^2}$

so you have

$\displaystyle2\int\frac{3t^2+2t+3}{(1+t)(1+t^2)}\,\mathrm dt=4\int\frac{\mathrm dt}{1+t}+2\int\frac{\mathrm dt}{1+t^2}+\int\frac{2t}{1+t^2}\,\mathrm dt$

which are all standard integrals. You end up with

$\displaystyle\int\mathrm dx+4\int\frac{\mathrm dt}{1+t}+2\int\frac{\mathrm dt}{1+t^2}+\int\frac{2t}{1+t^2}\,\mathrm dt$
$=x+4\ln|1+t|+2\arctan t+\ln(1+t^2)+C$
$=x+4\ln\left|1+\tan\dfrac x2\right|+2\arctan\left(\arctan\dfrac x2\right)+\ln\left(1+\tan^2\dfrac x2\right)+C$
$=2x+4\ln\left|1+\tan\dfrac x2\right|+\ln\left(\sec^2\dfrac x2\right)+C$

To try to get the terms to match up with the available answers, let's add and subtract $\ln\left|1+\tan\dfrac x2\right|$ to get

$2x+5\ln\left|1+\tan\dfrac x2\right|+\ln\left(\sec^2\dfrac x2\right)-\ln\left|1+\tan\dfrac x2\right|+C$
$2x+5\ln\left|1+\tan\dfrac x2\right|+\ln\left|\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}\right|+C$

which suggests A may be the answer. To make sure this is the case, show that

$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\sin x+\cos x+1$

You have

$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac1{\cos^2\dfrac x2+\sin\dfrac x2\cos\dfrac x2}$
$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac1{\dfrac{1+\cos x}2+\dfrac{\sin x}2}$
$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac2{\cos x+\sin x+1}$

So in the corresponding term of the antiderivative, you get

$\ln\left|\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}\right|=\ln\left|\dfrac2{\cos x+\sin x+1}\right|$
$=\ln2-\ln|\cos x+\sin x+1|$

The $\ln2$ term gets absorbed into the general constant, and so the antiderivative is indeed given by A,

$\displaystyle\int\frac{\cos x+3\sin x+7}{\cos x+\sin x+1}\,\mathrm dx=2x+5\ln\left|1+\tan\dfrac x2\right|-\ln|\cos x+\sin x+1|+C$

5 0

2 years ago