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melamori03 [73]
3 years ago
10

A 20-ounce candle is expected to burn for 60 hours. A 12-ounce candle is expected to burn for 36 hours. Assuming the variables a

re directly related, how many hours would a 9-ounce candle be expected to burn? 18
Mathematics
2 answers:
ANTONII [103]3 years ago
5 0
Because the ratio for both candles is 20/60 or 1/3 then take the number of ounces the candle has and divide by 1/3. In expression: 9 / 1/3 => 9 * 3 = 18
alina1380 [7]3 years ago
3 0

Answer:  A 9-ounce candle is expected to burn for 27 hours.

Step-by-step explanation:  Given that a 20-ounce candle is expected to burn for 60 hours and a 12-ounce candle is expected to burn for 36 hours.

If the variables are directly related, we are to find the number of hours that a 9-ounce candle is expected to burn.

Let, x represents the number of ounces of the candle and y represents the corresponding number of hours for which it burns.

Then, since the variables are directly related, the graph will be a straight line.

And, the two points (x, y) = (20, 60) and (12, 36) lies on the line.

So, the slope of the line will be

m=\dfrac{36-60}{12-20}\\\\\\\Rightarrow m=\dfrac{-24}{-8}\\\\\Rightarrow m=3.

Therefore, the equation of the line is

y-36=m(x-12)\\\\\Rightarrow y-36=3(x-12)\\\\\Rightarrow y=3x.

So, if x = 9, then

y=3\times9=27.

Thus, a 9-ounce candle is expected to burn for 27 hours.

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For what values of x is f(x) = |x + 1| differentiable? I'm struggling my butt off for this course
pav-90 [236]

By definition of absolute value, you have

f(x) = |x+1| = \begin{cases}x+1&\text{if }x+1\ge0 \\ -(x+1)&\text{if }x+1

or more simply,

f(x) = \begin{cases}x+1&\text{if }x\ge-1\\-x-1&\text{if }x

On their own, each piece is differentiable over their respective domains, except at the point where they split off.

For <em>x</em> > -1, we have

(<em>x</em> + 1)<em>'</em> = 1

while for <em>x</em> < -1,

(-<em>x</em> - 1)<em>'</em> = -1

More concisely,

f'(x) = \begin{cases}1&\text{if }x>-1\\-1&\text{if }x

Note the strict inequalities in the definition of <em>f '(x)</em>.

In order for <em>f(x)</em> to be differentiable at <em>x</em> = -1, the derivative <em>f '(x)</em> must be continuous at <em>x</em> = -1. But this is not the case, because the limits from either side of <em>x</em> = -1 for the derivative do not match:

\displaystyle \lim_{x\to-1^-}f'(x) = \lim_{x\to-1}(-1) = -1

\displaystyle \lim_{x\to-1^+}f'(x) = \lim_{x\to-1}1 = 1

All this to say that <em>f(x)</em> is differentiable everywhere on its domain, <em>except</em> at the point <em>x</em> = -1.

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3 years ago
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rusak2 [61]
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3 years ago
The difference between two numbers is 9. The first number plus twice the other number is 27. Find the two numbers.​
Ad libitum [116K]
X - y = 9
x + 2y = 27
You can subtract the top equation from the bottom
3y = 18
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The two numbers are 15 and 6
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3 years ago
Find the critical numbers of the function f(x) = x6(x − 1)5 what does the first derivative test tell you that the second derivat
patriot [66]
The given function is f(x) = x⁶(x-1)⁵

The first derivative is
f'(x) = 6x⁵(x-1)⁵ + 5x⁶(x-1)⁴
       = x⁵(x-1)⁴(6x - 6 + 5x)
       = x⁵(x-1)⁴(11x - 6)
The critical values are the zeros of f'(x). They are
x = 0, 6/11, and 1.
The critical values indicate that turning points exist for f(x) at the critical points. However, we do not know the nature of the turning points.

Write the first derivative in the form f'(x) = (x-1)⁴(11x⁶ - 6x⁵).
The second derivative is
f''(x) = 4(x-1)³(11x⁶ - 6x⁵) + (x-1)⁴(66x⁵ - 30x⁴)
        = (x-1)³(44x⁶ - 24x⁵ + 66x⁶ - 30x⁵ - 66x⁵ + 30x⁴)
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The sign of f''(x) at the critical values tell us the nature of the turning point.

f''(0) = 0, therefore a point of inflection exists at x = 0.
f''(6/11) > 0, therefore a local minumum exists at x = 6/11.
f''(1) = 0, therefore a point of inflection exists at x=1.

The graphs shown below confirm these results.

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3 years ago
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