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drek231 [11]
3 years ago
12

I don’t get this question so can someone please help.!

Mathematics
1 answer:
Svetllana [295]3 years ago
3 0
The answer is B,
see explanation in the attached picture

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Help with number 1!! And also 1 week= 5 days in this exercise okay!
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Each student collects $1 per week.
There are 24 students, so combined, they collect $24 per week.
The time irt takes to collect $1500 is

$1500/$24 = 62.5

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If f(x)= 2(3x ) +1, what is the value of f(2)? Students the x is an exponent so the question should read 2 ( 3 raised to the pow
vladimir2022 [97]

Answer:

f(x) = 2(3^x) + 1

f(2) = 2(3^2) + 1

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3(x-6)-8x = -2+5(2x+1)
Len [333]
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8 0
3 years ago
Let <img src="https://tex.z-dn.net/?f=f%28x%2Cy%29%20%3D%20%5Cfrac%7Bx%5E%7B2%7D%20y%7D%7Bx%5E%7B4%7D%2By%5E%7B2%7D%20%7D" id="T
stiv31 [10]

The correct answer is C.

The limit in A does exist:

\displaystyle \lim_{x\to0} f(x,0) = \lim_{x\to0} \frac0{x^4} = 0

The limit in B also exists: for any k\in\Bbb R,

\displaystyle \lim_{x\to0} f(x,kx) = \lim_{x\to0} \frac{kx^3}{x^4+k^2x^2} = \lim_{x\to0}\frac{kx}{x^2+k^2} = 0

But this alone does not prove the 2D limit exists. y=kx only captures all the paths through the origin that are straight lines.

The limit in C also exists, but it's not the same as either of the limits along the paths used in A and B.

\displaystyle \lim_{x\to0} f(x,x^2) = \lim_{x\to0} \frac{x^4}{2x^4} = \frac12

That this value is non-zero tells us the original limit does not exist.

The claim in D is generally not correct. That f(0,0) is undefined does not automatically mean the limit doesn't exist. A simpler example:

\displaystyle \lim_{x\to0} \frac{x}{x} = \lim_{x\to0} 1 = 1

yet \frac00 is undefined.

3 0
1 year ago
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