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Svetlanka [38]
3 years ago
5

The Constant of Proportionality from the following equation is: y= 7/9x

Mathematics
2 answers:
Ilia_Sergeevich [38]3 years ago
5 0

Answer:

k = \frac{7}{9}

Step-by-step explanation:

The equation of proportionality is

y = kx ← k is the constant of proportionality

y = \frac{7}{9} x ← is in this form

with k = \frac{7}{9}

Andrej [43]3 years ago
3 0

Answer:

\frac{7}{9}

Step-by-step explanation:

A direct variation function is in this general form:

y = kx\\\rule{150}{0.5}\\k -\text{ Constant of Proportionality}

Since \frac{7}{9} is in 'k's place, it is the Constant of Proportionality.

Hope this helps you.

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If ƒ(x ) = 3x + 1 and ƒ -1 = , then ƒ -1(7) =
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F(x) = 3x + 1

let f(x) = y

y = 3x + 1

y - 1 = 3x

3x = y - 1

x = (y - 1)/3

From y = f(x),  x = f⁻¹(y) 

<span>x = (y - 1)/3
</span>
 f⁻¹(y)  = (y - 1)/3

f⁻¹(7)  = <span>(7 - 1)/3 = 6/3 = 2

</span>f⁻¹(7)<span> = 2</span>
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How many integers between 10000 and 99999, inclusive, are divisible by 3 or<br> 5 or 7?
Yuki888 [10]

Answer: Hence, there are approximately 48884 integers are divisible by 3 or 5 or 7.

Step-by-step explanation:

Since we have given that

Integers between 10000 and 99999 = 99999-10000+1=90000

n( divisible by 3) = \dfrac{90000}{3}=30000

n( divisible by 5) = \dfrac{90000}{5}=18000

n( divisible by 7) = \dfrac{90000}{7}=12857.14

n( divisible by 3 and 5) = n(3∩5)=\dfrac{90000}{15}=6000

n( divisible by 5 and 7) = n(5∩7) = \dfrac{90000}{35}=2571.42

n( divisible by 3 and 7) = n(3∩7) = \dfrac{90000}{21}=4285.71

n( divisible by 3,5 and 7) = n(3∩5∩7) = \dfrac{90000}{105}=857.14

As we know the formula,

n(3∪5∪7)=n(3)+n(5)+n(7)-n(3∩5)-n(5∩7)-n(3∩7)+n(3∩5∩7)

=30000+18000+12857.14-6000-2571.42-4258.71+857.14\\\\=48884.15

Hence, there are approximately 48884 integers are divisible by 3 or 5 or 7.

5 0
4 years ago
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