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Ilia_Sergeevich [38]
3 years ago
11

Determine all critical points for the following function. f(x)=x2−14x+1

Mathematics
1 answer:
melomori [17]3 years ago
4 0

9514 1404 393

Answer:

  (x, y) = (7, -48)

Step-by-step explanation:

A critical point is where the slope is zero or undefined. A polynomial function will never have any points where the slope is undefined.

The slope is zero were f'(x) = 0.

  f'(x) = 2x -14

  0 = 2x -14

  0 = x -7

  7 = x

  f(7) = 49 -98 +1 = -48

The critical point is (7, -48).

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Hi, I don’t understand this question
meriva

Given that

z₁ = 15 (cos(90°) + i sin(90°))

z₂ = 3 (cos(10°) + i sin(80°))

we get the quotient z₁/z₂ by dividing the moduli and subtracting the arguments:

z₁/z₂ = 15/3 (cos(90° - 10°) + i sin(90° - 10°))

z₁/z₂ = 5 (cos(80°) + i sin(80°))

so that z₁ is scaled by a factor of 1/3 and is rotated 10° clockwise.

4 0
2 years ago
How many bits are required to represent the decimal numbers in the range from 0 to 999 in straight binary code?
ad-work [718]
Note that powers of 2 can be written in binary as

2^0=1_2
2^1=10_2
2^2=100_2

and so on. Observe that n+1 digits are required to represent the n-th power of 2 in binary.

Also observe that

\log_2(2^n)=n\log_22=n

so we need only add 1 to the logarithm to find the number of binary digits needed to represent powers of 2. For any other number (non-power-of-2), we would need to round down the logarithm to the nearest integer, since for example,

2_{10}=10_2\iff\log_2(2^1)=\log_22=1
3_{10}=11_2\iff\log_23=1+(\text{some number between 0 and 1})
4_{10}=100_2\iff\log_24=2

That is, both 2 and 3 require only two binary digits, so we don't care about the decimal part of \log_23. We only need the integer part, \lfloor\log_23\rfloor, then we add 1.

Now, 2^9=512, and 999 falls between these consecutive powers of 2. That means

\log_2999=9+\text{(some number between 0 and 1})

which means 999 requires \lfloor\log_2999\rfloor+1=9+1=10 binary digits.

Your question seems to ask how many binary digits in total you need to represent all of the numbers 0-999. That would depend on how you encode numbers that requires less than 10 digits, like 1. Do you simply write 1_2? Or do you pad this number with 0s to get 10 digits, i.e. 0000000001_2? In the latter case, the answer is obvious; 1000\times10=10^4 total binary digits are needed.

In the latter case, there's a bit more work involved, but really it's just a matter of finding how many number lie between successive powers of 2. For instance, 0 and 1 both require one digit, 2 and 3 require two, while 4-7 require three, while 8-15 require four, and so on.
8 0
3 years ago
5x-7=118 solve for the variable x
myrzilka [38]
X=25 ...... ................
5 0
3 years ago
Read 2 more answers
Which number is a perfect cube? 5, 100, 125, 150​
KATRIN_1 [288]

Answer:

125

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
How many times smaller is 2 × 10^-10 than 4 × 10^-5
natta225 [31]

Answer:

200,000

Step-by-step explanation:

To find how many times smaller is smaller number than greater number, divide greater number by smaller number.

Greater number =4\times 10^{-5}

Smaller number =2\times 10^{-10}

Then

\dfrac{\text{Greater number}}{\text{Smaller number}}\\ \\=\dfrac{4\times 10^{-5}}{2\times 10^{-10}}\\ \\=\dfrac{4}{2}\times \dfrac{10^{-5}}{10^{-10}}\\ \\=2\times \dfrac{1}{10^{-5}}\\ \\=2\times 10^5\\ \\=2\times 100,000\\ \\=200,000

7 0
3 years ago
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