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Rom4ik [11]
3 years ago
14

If a is an arbitrary nonzero constant, what happens to a/b as b approaches 0

Mathematics
1 answer:
Stolb23 [73]3 years ago
7 0

It depends on how b approaches 0

If b is positive and gets closer to zero, then we say b is approaching 0 from the right, or from the positive side. Let's say a = 1. The equation a/b turns into 1/b. Looking at a table of values, 1/b will steadily increase without bound as positive b values get closer to 0.

On the other side, if b is negative and gets closer to zero, then 1/b will be negative and those negative values will decrease without bound. So 1/b approaches negative infinity if we approach 0 on the left (or negative) side.

The graph of y = 1/x shows this. See the diagram below. Note the vertical asymptote at x = 0. The portion to the right of it has the curve go upward to positive infinity as x approaches 0. The curve to the left goes down to negative infinity as x approaches 0.

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120 inches

Step-by-step explanation:

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Choose a method you prefer to solve for a system of linear equations.<br> 2x + 4y = -4<br> 2x-y=6
Artyom0805 [142]

Answer:

Elimination

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Step-by-step explanation:

Elimination

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3 years ago
What is the circumference of this circle? (Use C = <img src="https://tex.z-dn.net/?f=%5Cpi" id="TexFormula1" title="\pi" alt="\p
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Read 2 more answers
Suppose you choose a team of two people from a group of n &gt; 1 people, and your opponent does the same (your choices are allow
jonny [76]

Answer:

The number of possible choices of my team and the opponents team is

 \left\begin{array}{ccc}n-1\\E\\n=1\end{array}\right     i^{3}

Step-by-step explanation:

selecting the first team from n people we have \left(\begin{array}{ccc}n\\1\\\end{array}\right)  = n possibility and choosing second team from the rest of n-1 people we have \left(\begin{array}{ccc}n-1\\1\\\end{array}\right) = n-1

As { A, B} = {B , A}

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Since our choices are allowed to overlap, the second team is \frac{n(n-1)}{2}

Possibility of choosing both teams will be

\frac{n(n-1)}{2}  *  \frac{n(n-1)}{2}  \\\\= [\frac{n(n-1)}{2}] ^{2}

We now have the formula

1³ + 2³ + ........... + n³ =[\frac{n(n+1)}{2}] ^{2}

1³ + 2³ + ............ + (n-1)³ = [x^{2} \frac{n(n-1)}{2}] ^{2}

=\left[\begin{array}{ccc}n-1\\E\\i=1\end{array}\right] =   [\frac{n(n-1)}{2}]^{3}

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