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Allushta [10]
2 years ago
6

find the distance of CD if the coordinates of C are (-4,11) and D are ( -3,4) Express your answer in simplest radical form.

Mathematics
1 answer:
Aloiza [94]2 years ago
4 0

Step-by-step explanation:

it's not clear but I tried

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A student wants to save at least $5000 for college. She begins with a savings account set up by her parents containing $2,500. E
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Answer: She deposits 50 dollars per month right? But as known her parents deposited her 2,500. So, it will take her 50 months to complete the amount of  5000 dollars needed for college.

Step-by-step explanation: Im not good with equations but here  we go.

First: AS we know the account already had 2,500, and every month she deposits 50$.

What we can do is to divide 2,500 divided by 50 which will be 50.

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How can finding the slopes of MN and AB help in proving that MN is the midsegment of triangle ABC?
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Mid-segment and bottom side of the triangle is always parallel to each other, and parallel lines must have same slopes, so if their slopes are equal, then MN would be mid-segment of side AB in Triangle ABC

Hope this helps!
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4 years ago
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What is the measure of one angle of a regular 12-gon
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(N-2) * 180 so is 1800
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4 years ago
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6.6÷2.2+(3.2+4)=<br> Use pemdas
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3.2+4=7.2
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8 0
3 years ago
Each of these extreme value problems has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to f
tino4ka555 [31]

Answer:

The minimum value of the given function is f(0) = 0

Step-by-step explanation:

Explanation:-

Extreme value :-  f(a, b) is said to be an extreme value of given function 'f' , if it is a maximum or minimum value.

i) the necessary and sufficient condition for f(x)  to have a maximum or minimum at given point.

ii)  find first derivative f^{l} (x) and equating zero

iii) solve and find 'x' values

iv) Find second derivative f^{ll}(x) >0 then find the minimum value at x=a

v) Find second derivative f^{ll}(x) then find the maximum value at x=a

Problem:-

Given function is f(x) = log ( x^2 +1)

<u>step1:</u>- find first derivative f^{l} (x) and equating zero

  f^{l}(x) = \frac{1}{x^2+1} \frac{d}{dx}(x^2+1)

f^{l}(x) = \frac{1}{x^2+1} (2x)  ……………(1)

f^{l}(x) = \frac{1}{x^2+1} (2x)=0

the point is x=0

<u>step2:-</u>

Again differentiating with respective to 'x', we get

f^{ll}(x)=\frac{x^2+1(2)-2x(2x)}{(x^2+1)^2}

on simplification , we get

f^{ll}(x) = \frac{-2x^2+2}{(x^2+1)^2}

put x= 0 we get f^{ll}(0) = \frac{2}{(1)^2}   > 0

f^{ll}(x) >0 then find the minimum value at x=0

<u>Final answer</u>:-

The minimum value of the given function is f(0) = 0

5 0
4 years ago
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