What is 2 6/7 as an improper fraction

PLEASE HELP ME FAST!!!!

Korolek [52]

Answer:

ctrl+k

Step-by-step explanation:

it allows you to search a url or paste a url in your assignment

7 0

3 years ago

Sin(A+B) sin(A-B) /sin^A Cos^B=1-cot^A Tan^B

telo118 [61]

In order to prove

$\dfrac{\sin(x+y)\sin(x-y)}{\sin^2(x)\cos^2(x)}=1-\cot^2(x)\tan^2(y)$

Let's write both sides in terms of $\sin(x),\ \sin^2(x),\ \cos(x),\ \cos^2(x)$ only.

Let's start with the left hand side: we can use the formula for sum and subtraction of the sine to write

$\sin(x+y)=\cos(y)\sin(x)+\cos(x)\sin(y)$

and

$\sin(x-y)=\cos(y)\sin(x)-\cos(x)\sin(y)$

So, their multiplication is

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

So, the left hand side simplifies to

$\dfrac{\cos^2(y)\sin^2(x)-\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}$

Now, on with the right hand side. We have

$1-\cot^2(x)\tan^2(y)=1-\dfrac{\cos^2(x)}{\sin^2(x)}\cdot\dfrac{\sin^2(y)}{\cos^2(y)} = 1-\dfrac{\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}$

Now simply make this expression one fraction:

$1-\dfrac{\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}=\dfrac{\sin^2(x)\cos^2(y)-\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}$

And as you can see, the two sides are equal.

6 0

4 years ago

How many solutions does the equation 3x − 7 = 4 + 6 + 4x have

Sunny_sXe [5.5K]

Answer:

The given equation 3x − 7 = 4 + 6 + 4x has one solution.

Step-by-step explanation:

Given : Equation 3x − 7 = 4 + 6 + 4x

We have to find the number of solutions of the given equation.

Consider the given equation 3x − 7 = 4 + 6 + 4x

Simplify, we have,

3x − 7 = 10 + 4x

Now subtract 3x both side, we hvae,

-7 = 10 + 4x - 3x

Simplify, we have,

- 7 = 10 + x

Subtract 10 both side, we have,

-7 - 10 = x

x = - 17

Thus, The given equation 3x − 7 = 4 + 6 + 4x has one solution.

8 0

4 years ago

Read 2 more answers

QUICKKK PLEASE HELP!!!! Q:43c +10d=?????

vlabodo [156]

Answer:

53d

Step-by-step explanation:

6 0

3 years ago

Determine how many different computer passwords are possible if (a) digits and letter so can be repeated and (b) digits and lett

Lerok [7]

Answer:

a) 1,188,137,600 different passwords.

b) 710,424,000 different passwords.

Step-by-step explanation:

We have a code of 2 digits followed by 5 letters.

First, the total number of digits is 10

The total number of letters is 26.

Then:

a) Digits and letters can be repeated.

Here we need to count the number of options for each selection.

For the first digit, we have 10 options.

For the second digit, we have 10 options.

For the first letter, we have 26 options.

For the second letter, we have 26 options.

For the third letter, we have 26 options.

For the fourth letter, we have 26 options.

For the fifth letter, we have 26 options.

The total number of combinations will be equal to the product of the number of options. We get:

Combinations = 10*10*26*26*26*26*26 = (10^2)*(26^5) = 1,188,137,600

This means that we have 1,188,137,600 different possible passwords.

b) Digits and letters can not be repeated.

We start in the same way as above:

For the first digit, we have 10 options.

For the second digit, we have 9 options, because one is already used.

For the first letter, we have 26 options.

For the second letter, we have 25 options, because one letter is already used.

For the third letter, we have 24 options, because 2 letters are already used.

For the fourth letter, we have 23 options, because 3 letters are already used.

For the fifth letter, we have 22 options, because 4 letters are already used.

Then the total number of combinations is:

Combinations = 10*9*26*25*24*23*22 = 710,424,000

So if we can not repeat digits nor letters, we can make 710,424,000 different passwords,

8 0

3 years ago