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Ierofanga [76]
3 years ago
9

I need help with numbers 6 and 7

Mathematics
2 answers:
IgorLugansk [536]3 years ago
8 0

Answer:

we would need the ratios... please comment the ratios and maybe I can help

Step-by-step explanation:

ella [17]3 years ago
4 0
I need to see more of the picture
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What number means the same as 9,000,000,000 + 900,000,000, + 90,000,000 +9,000,000, + 900,000 + 9,000 +9 PLEASE DO IT QUICKLY
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9,999,909,009

Step-by-step explanation:

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Is the expression 7 (6x-8y) equal to 42x-56y
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Please help me i beg of u due today
Vlad1618 [11]

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4 0
3 years ago
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Solve the equation log^7b>2
ahrayia [7]

Answer:

b > 12.71

Step-by-step explanation:

Here in this question we have to solve the inequality equation given for b (An unknown variable}

Now, the equation is (\log b)^{7} > 2

⇒ \log b > 2^{\frac{1}{7} }

We are taking the base of the log here is 10.

So, 10^{(\log b)} > 10^{2^{\frac{1}{7} } } {Since 10^{\log x} =x and e^{\ln x} = x}

⇒ b > 12.71 (Approximate) (Answer)

3 0
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)
        = (x-1)³(110x⁶ - 120x⁵ + 30x⁴)
        = 10x⁴(x-1)³(11x² - 12x + 3)
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.

8 0
3 years ago
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