Answer:
2A9
Step-by-step explanation:
end of the diving board is the diver when he hits the water?
Answer:
The diver will be 8 feet from the end of the board when he hits the water.
Step-by-step explanation:
The diver hits the water when y = 0.
To find the distance, we have to find the values of x when y = 0.
Solving a quadratic equation:
Given a second order polynomial expressed by the following equation:
.
This polynomial has roots such that , given by the following formulas:
In this problem, we have that:
So
Then
It is a horizontal distance, so the answer is a positive value.
The diver will be 8 feet from the end of the board when he hits the water.
Step-by-step explanation:
1350÷600
=2.25
√2.25=1.5
Answer:
This is correct.
Step-by-step explanation:
Answer:
6y +52
Step-by-step explanation:
6(2y + 8) - 2(3y - 2)
Distribute
12y + 48 - 6y +4
Combine like terms
6y +52
Answer:
More number of words that can be made:
Please refer to below proof.
Step-by-step explanation:
Given that:
The number of binary code words that can be made:
where n is the length of binary numbers.
Binary numbers means 2 possibilities either 0 or 1.
Here, suppose if we have 5 as the length of binary number.
And there are 2 possibilities for each digit.
So, total number of possibilities will be
If the length of binary number is 2.
The total words possible are .
These numbers are:
{00, 01, 10, 11}
If the length of binary number is 3. (increasing the 'n' by 1)
The total words possible are .
These words are:
{000, 001, 010, 100, 011, 101, 110, 111}
So, number of More binary words = 8 - 4 = 4 or or .
So, the answer is .
Let us try to prove in generic terms:
Increasing the n by 1:
Number of more words made by increasing n by 1:
Hence, proved that:
More number of words that can be made:
When the length of a binary code word increases from n to n+1, the number of additional binary code words is equal to the number of binary code words of length n, which is .
When the length is increased from n to n+1, the number of binary code words of length n+1 is equal to the number of binary code words of length n multiplied by 2. This is because for each binary code word of length n, we can append a 0 or a 1 to create two new binary code words of length n+1. Therefore, the number of additional binary code words is equal to the number of binary code words of length n, which is.
#SPJ3
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Explanation:
The exponent rules are
The expression on the right hand side simplifies to c^(-1). We add the exponents to get this. So we're using rule 1 mentioned above.
We'll keep this in mind for later.
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On the left hand side (c^5)^4 becomes c^(20) because we multiply the exponents (rule 3)
Then (c^20)/(c^x) becomes c^(20-x). We subtract exponents here (rule 2).
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After using those exponent rules, the original equation turns into
c^(20-x) = c^(-1)
The bases are both c, so the exponents must be equal
20-x = -1
-x = -1-20
-x = -21
x = 21