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joja [24]
3 years ago
5

Find the exact value of each expression. sin75 degrees and cos(-75 degrees).​

Mathematics
1 answer:
valkas [14]3 years ago
6 0

Answer:

0.965925826

0.258819045

Step-by-step explanation:

→If you were to calculate sin(75), you would have a total of:

0.965925826

→If you were to calculate cos(-75), you would have a total of:

0.258819045

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If f(x)=x-6 and g(x)=x^1/2(x+3) what is f(x) times g(x
dimaraw [331]
This  is (x - 6)*x^1/2(x + 3)

= x^1/2 (x^2 - 3x - 18)

= x^(5/2) - 3x^(3/2) - 18x^(1/2)
7 0
3 years ago
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Solve.<br> x² – 8x - 12 = 36
professor190 [17]

sorry answered the wrong one..

7 0
3 years ago
Set D is the set of positive two-digit even numbers less than 25 that do not contain the digit 0.
gavmur [86]

The required set of positive two-digit even numbers less than 25 that do not contain the digit 0 is 12,14,16,18,22,24.

<h3>What is integer?</h3>

An integer is a whole number (not a fractional number) that can be positive, negative, or zero. Zero is not a fraction or decimal of any number. It is neither positive nor negative.

Given:

Set D is the set of positive two-digit even numbers less than 25 that do not contain the digit 0.

According to given question we have

2-digit numbers are the numbers that have two digits and they start from the number 10 and end on the number 99.

Starting at 12 going up the two digit numbers even numbers less than 25 that do not contain the digit 0.

12,14,16,18,22,24

Therefore, the required set of positive two-digit even numbers less than 25 that do not contain the digit 0 is 12,14,16,18,22,24.

Learn more details about integer here:

brainly.com/question/15276410

#SPJ1

4 0
1 year ago
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
3 years ago
Recall that a 6-bit string is a bit strings of length 6, and a bit string of weight 3, say, is one with exactly three 1's. How m
strojnjashka [21]

Answer:

1.. Total number of 6 bit strings is 64

2. Number of 6-bit strings with weight of 0 is 1

3. Number of 6-bit strings with weight of 1 is 6

4. Number of 6-bit strings with weight of 3 is 20

5. Number of 6-bit strings with weight of 5 is 6

6. Number of 6-bit strings with weight of 6 is 1

7. Number of 6-bit strings with weight of 7 is 0

Step-by-step explanation:

A bit string is a string that contains 0 and 1 only

1. Total number of 6 bit strings is 2^6 = 64

2. Number of 6 bit strings with weight 0 is 1

Explanation

Weight 0 means a string with no occurrence of 1

Here, we are only interested in occurrence and not order of occurrence

We apply combination formula for this

nCr = n!/(n-r)!r!

n = 6 and r = 0 i.e. no occurrence of 1

6C0 = 6!/(6-0)!0!

6C0 = 6!/6!0!

6C0 = 1

Hence, the number of string with weight 0 (i.e. no occurrence of 1 ) is 1

3. Number of string with weight 1 is 6

Explanation

Weight 0 means a string with exactly 1 occurrence of '1'

Here, we are only interested in occurrence and not order of occurrence

We apply combination formula for this

nCr = n!/(n-r)!r!

n = 6 and r = 1

6C1 = 6!/(6-1)!1!

6C1 = 6!/5!1!

6C1 = 6

Hence, the number of string with weight 6

4. Number of string with weight 3 is 20

Explanation

n = 6 and r = 3

6C3 = 6!/(6-3)!3!

6C3 = 6!/3!3!

6C3 = 20

Hence, the number of string with weight 3 is 20

5. Number of string with weight 5 is 6

Explanation

n = 6 and r = 5

6C5 = 6!/(6-5)!5!

6C5 = 6!/1!5!

6C5 = 6

Hence, the number of string with weight 5 is 6

6. Number of string with weight 6 is 1

Explanation

n = 6 and r = 6

6C6 = 6!/(6-6)!6!

6C6 = 6!/0!6!

6C6 = 1

Hence, the number of string with weight 6 is 1

7. Number of string with weight 7 is 0

Weight of 7 means that a string that has 7 occurrence of 1

The total length of a 6 bit is 6

Since 6 is less than 7, there's no way a bit of weight 7 can occur.

So, the right answer for this is 0.

8 0
3 years ago
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