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slamgirl [31]
3 years ago
11

Find two numbers whose sum is 25 and the sum of whose reciprocals is 1/6

Mathematics
1 answer:
Katen [24]3 years ago
5 0

Answer:

10 and 15

Step-by-step explanation:

Let 'x' and 'y' are the numbers we need to find.

x + y = 25 (two numbers whose sum is 25)

(1/x) + (1/y) = 1/6 (the sum of whose reciprocals is 1/6)

The solutions of the this system of equations are the numbers we need to find.

x = 25 - y

1/(25 - y) + 1/y = 1/6 multiply both sides by 6(25-y)y

6y + 6(25-y) = (25-y)y

6y + 150 - 6y = 25y - (y^2)

y^2 - 25y + 150 = 0 quadratic equation has 2 solutions

y1 = 15

y2 = 10

Thus we have :

First solution: for y = 15, x = 25 - 15 = 10

Second solution: for y = 10, x = 25 - 10 = 15

The first and the second solution are in fact the same one solution we are looking for: the two numbers are 10 and 15 (since the combination 10 and 15 is the same as 15 and 10).

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5 inches

Step-by-step explanation:

See attachment for explanation.

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3 years ago
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When we approach limits, we are finding values that are infinitesimally approaching this x-value. Essentially, we consider the approximate location that this root or limit appears. This is essential when it comes to taking Calculus, and finding the limit or rate of change of a function.

When we are attempting limits questions, there are several tests we attempt first.

1. Evaluate the limit by substituting the value of the x-value as it approaches the value (direct evaluation of a limit)
2. Rearrangement of the function, such that we can evaluate the limit.
3. (TRIGONOMETRIC PROPERTIES)
\lim_{x \to 0} (\frac{sinx}{x}) = 1
\lim_{x \to 0} (\frac{tanx}{x}) = 1
4. Using L'Hopital's Rule for indeterminate limits, such as 0/0, -infinity/infinity, or infinity/infinity.

For example:

1) \lim_{x \to 0}\frac{\sqrt{x} - 5}{x - 25}

We can do this using the first and second method.
<em>Method 1: Direct evaluation:</em>

Substitute x = 0 to the function.
\frac{\sqrt{0} - 5}{0 - 25}
= \frac{-5}{-25}
= \frac{1}{5}

<em>Method 2: Rearranging the function
</em>

We can see that x - 25 can be rewritten as: (√x - 5)(√x + 5)
By rewriting it in this form, the top will cancel with the bottom easily, and our limit comes out the same.

\lim_{x \to 0}\frac{(\sqrt{x} - 5)}{(\sqrt{x} - 5)(\sqrt{x} + 5)}
= \lim_{x \to 0}\frac{1}{(\sqrt{x} + 5)}}
= \frac{1}{5}

Every example works exactly the same way, and by remembering these criteria, every limit question should come out pretty naturally.
8 0
3 years ago
A rectangle with a perimeter of 19m^2+2m-10 and a width of m^2 write an expression for the lenght
expeople1 [14]

Answer:

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Step-by-step explanation:

Since perimeter, P=19m^{2}+ 2m -10 and we know that P=2(l+w) then P=2(l+m^{2})=2l+2m^{2}

The length will be 2l=19m^{2}+ 2m -10-2m^{2}=17m^{2}+ 2m -10

Now l=8.5m^{2}+ m -5

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