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enyata [817]
2 years ago
7

0.43 (3 repeating) as a fraction

Mathematics
2 answers:
valina [46]2 years ago
8 0

If <em>x</em> = 0.4333…, then

10<em>x</em> = 4.333…

100<em>x</em> = 43.333…

so that

100<em>x</em> - 10<em>x</em> = 43.333… - 4.333…

90<em>x</em> = 39

<em>x</em> = 39/90 = 13/30

DiKsa [7]2 years ago
7 0

Your answer is 13/30

I hope this helps and I hope you have a wonderful day/night, may God bless!

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9(k-2)=-6 what is K the oz
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Answer:k=0.67

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$12 per 45 minutes is how much an hour? (I will give good rating to anyone who answers)
uysha [10]

Answer:

$16

Step-by-step explanation:

1. Simplify 12/45 which is now 4/15. Now we know $4 is made every 15 minutes.

2. Divide into 60(minutes) by the the 15. So 60÷15=4. Now we the number of times 15 can be divided into 60.

3. Multiply the ratio 4/15 by 4/4 to see how much is made in an hour. 4x4/15×4=16/60.

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3 years ago
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Use the distributive property to write an expression that is equivalent to:<br> 3(x + 5)
lisov135 [29]

Answer: 3x + 15

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In order to use the distributive property, you simply multiply the number on the outside by both the number inside the parenthesis

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3 years ago
Find the critical numbers and the intervals on which the function f(x)=9x5−3x3+6f(x)=9x5−3x3+6 is increasing or decreasing. Use
Ierofanga [76]

Answer:

x = (0, -√(3/5), √(3/5)) are the critical points.

Local maximum at √(3/5), and local minimum is at -√(3/5).

The function is increasing in the interval

(−∞, -√(3/5)) U (√(3/5), ∞)

And decreasing in the interval

(-√(3/5), 0) U (0, √(3/5))

Step-by-step explanation:

Given the function

f(x) = 9x^5 - 3x³ + 6

First of all, take the first derivative of this, to have

f'(x) = 45x^4 - 27x²

The critical point are the points where the first derivative vanishes, that is

f'(x) = 0

Now, solve the equation

45x^4 - 27x² = 0

9x²(5x² - 3) = 0

x = 0 twice

Or

5x² - 3 = 0

5x² = 3

x² = 3/5

x = ±√(3/5)

So, x = (0, -√(3/5), √(3/5)) are the critical points.

Local maximum is when f'(x) > 0, this is √(3/5) in this case,

and local minimum is when f'(x) < 0, this is -√(3/5) in this case.

Now, we need to test for the various intervals to determine where the function increases and decreases.

(−∞, -√(3/5)):

f'(-√(4/5)) = 45(-√(4/5))^4 - 27(-√(4/5))²

= 36/5 > 0. Increasing

(-√(3/5), 0):

f'(-√(2/5)) = 45(-√(2/5))^4 - 9(-√(2/5))²

= -18/5 < 0. Decreasing

(0, √(3/5)):

f'(√(1/5)) = 45(√(1/5))^4 - 9(√(1/5))² = -18/5 < 0. Decreasing

(√(3/5), ∞): f'(1) = 45(1)^4 - 9(1)² =

36 > 0. Increasing.

Therefore, the function is increasing in the interval

(−∞, -√(3/5)) U (√(3/5), ∞)

And decreasing in the interval

(-√(3/5), 0) U (0, √(3/5))

6 0
3 years ago
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