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Nataly_w [17]
3 years ago
8

37. A group of people share equally in a $20,000,000 lottery. Before

Mathematics
1 answer:
mart [117]3 years ago
5 0

Answer:

20 people

Step-by-step explanation:

You do 20,000,000 divided by 2 in a calculator, which is 10,000,000 then divide it by 500,000 to get 20

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4x+ 3y -2z=35, 2x-3y+z=-6, x+2y+3z=-5, solve for x,y and z
Mkey [24]
1. X = \frac{35}{4}+\frac{\left( -3\right) \ Y}{4}+\frac{z}{2}$$
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A circular picture is 8 inches in diameter.
velikii [3]

The formula of an area of a circle with radius r:


A=\pi r^2


Part A.

We have diameter of the circle d = 8 in. The diameter of a circle is equal two times length of a radius.

Therefore:


r=\dfrac{d}{2}\to r=\dfrac{8}{2}=4\ in


Calculate the area:


A=\pi\cdot4^2=16\pi\ in^2


Answer: C. 16π square inches.


Part 2.

Look at the picture.


Total area:


A=\pi\cdot6^2=36\pi\ in^2


Answer: C. 36π square inches

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3 years ago
Help!!!!! Math work!!!!!!
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Answer: y-intercept: (0,-6)
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Read 2 more answers
Order the sides of each triangle from shortest to longest.
Svetradugi [14.3K]

Answer:

A.

Step-by-step explanation:

the line that's PR is longer than OR

7 0
2 years ago
Does there exist a di↵erentiable function g : [0, 1] R such that g'(x) = f(x) for all x 2 [0, 1]? Justify your answer
agasfer [191]

Answer:

No; Because g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Step-by-step explanation:

Assuming:  the function is f(x)=x^{2} in [0,1]

And rewriting it for the sake of clarity:

Does there exist a differentiable function g : [0, 1] →R such that g'(x) = f(x) for all g(x)=x² ∈ [0, 1]? Justify your answer

1) A function is considered to be differentiable if, and only if  both derivatives (right and left ones) do exist and have the same value. In this case, for the Domain [0,1]:

g'(0)=g'(1)

2) Examining it, the Domain for this set is smaller than the Real Set, since it is [0,1]

The limit to the left

g(x)=x^{2}\\g'(x)=2x\\ g'(0)=2(0) \Rightarrow g'(0)=0

g(x)=x^{2}\\g'(x)=2x\\ g'(1)=2(1) \Rightarrow g'(1)=2

g'(x)=f(x) then g'(0)=f(0) and g'(1)=f(1)

3) Since g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Because this is the same as to calculate the limit from the left and right side, of g(x).

f'(c)=\lim_{x\rightarrow c}\left [\frac{f(b)-f(a)}{b-a} \right ]\\\\g'(0)=\lim_{x\rightarrow 0}\left [\frac{g(b)-g(a)}{b-a} \right ]\\\\g'(1)=\lim_{x\rightarrow 1}\left [\frac{g(b)-g(a)}{b-a} \right ]

This is what the Bilateral Theorem says:

\lim_{x\rightarrow c^{-}}f(x)=L\Leftrightarrow \lim_{x\rightarrow c^{+}}f(x)=L\:and\:\lim_{x\rightarrow c^{-}}f(x)=L

4 0
3 years ago
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