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4 years ago

How many two-digit positive integers can be formed from the digits 1, 5, 6, and 8 if no digit is repeated?

2 answers:

6 0

Write out all of the 2-digit positive integers formable from 1, 5, 6 and 8, without repetition of digits (e. g., no '22'):

15, 16, 18

Now write out the 2-digit pos. int. that begin with 5: 51, 56, 58.

And so on.

Remember that you can also form 2-dig. pos. int. beginning with 6 and beginning with 8.

Count how many 2-dig. pos. int. you have created. That's your answer.

Andreas93 [3]4 years ago

3 0

Answer:

The total number of ways to form a two number without repetition is 12.

Step-by-step explanation:

The given digits are 1, 5, 6 are 8. Total number of digits is 4.

We have to form a two digit number without repeating a digit. So, the total ways to choose a number for first place is 4.

Since repetition is not allowed, therefore the total ways to choose a number for second place is 3.

Total number of ways to form a two number without repetition is

$4\times 3=12$

Therefore the total number of ways to form a two number without repetition is 12.

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the difference of 2 times d minus 3= 6

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2 years ago

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Ierofanga [76]

7 is incorrect. The answer should be -4. Here's how I'd derive it:

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

With $\pi$ , we should expect $\cos x$ . If $\tan x=\dfrac8{15}$ , then

$\sec x=-\sqrt{1+\tan^2x}=-\dfrac{17}{15}\implies\cos x=-\dfrac{15}{17}$

Also,

$\pi$

so we should expect $\tan\dfrac x2$ and

$\tan\dfrac x2=-\sqrt{\dfrac{1-\cos x}{1+\cos x}}=-4$

You seem to be taking

$\tan\dfrac x2=\dfrac{1+\cos x}{-\sin x}$

but this is not an identity.

###

8 is incorrect. Just a silly mistake, you swapped the order of the terms in the numerator. It should be

$\dfrac{\sqrt6-\sqrt2}4$

###

9. Use the identity from (7). $\dfrac{7\pi}{12}$ lies in the second quadrant, so

$\tan\dfrac{7\pi}{12}=-\sqrt{\dfrac{1-\cos\frac{7\pi}6}{1+\cos\frac{7\pi}6}}=-\sqrt{7+4\sqrt3}=-2-\sqrt3$

###

12.

$\dfrac{\cot x-1}{1-\tan x}=\dfrac{\frac{\cos x}{\sin x}-1}{1-\frac{\sin x}{\cos x}}$

$=\dfrac{\cos x(\cos x-\sin x)}{\sin x(\cos x-\sin x)}$

$=\dfrac{\cos x}{\sin x}$

$=\dfrac{\csc x}{\sec x}$

###

13.

$\dfrac{1+\tan x}{\sin x+\cos x}=\dfrac{1+\frac{\sin x}{\cos x}}{\sin x+\cos x}$

$=\dfrac{\cos x+\sin x}{\cos x(\sin x+\cos x)}$

$=\dfrac1{\cos x}$

$=\sec x$

###

14.

$\sin2x(\cot x+\tan x)=2\sin x\cos x\left(\dfrac{\cos x}{\sin x}+\dfrac{\sin x}{\cos x}\right)$

$=2\sin x\cos x\dfrac{\cos^2x+\sin^2x}{\sin x\cos x}$

$=2$

###

15.

$\dfrac{1-\tan^2\theta}{1+\tan^2\theta}=\dfrac{1-\frac{\sin^2\theta}{\cos^2\theta}}{1+\frac{\sin^2\theta}{\cos^2\theta}}$

$=\dfrac{\cos^2\theta-\sin^2\theta}{\cos^2\theta+\sin^2\theta}$

$=\cos2\theta$

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3 years ago