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Anuta_ua [19.1K]
3 years ago
13

Consider the equation below.

Mathematics
1 answer:
svlad2 [7]3 years ago
6 0

Answer:

x = 4

Step-by-step explanation:

we are given the equation

<u>x/h + 1 = 2 </u>

we are also given a the<u> value of h = 4 </u>

<u>we are told to find x </u>

let's use the equations and substitute in what we have first

x/h + 1 = 2  ➔  x/4 + 1 = 2

now we can solve for x

x/4 + 1 = 2

lets begin by subtracting one from both sides

now we are left with

x/4 = 1

now do the inverse of division which is multiplication

so we multiply both sides by the denominator 4

now we have

x = 4  

Hope this helps!! :)

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Prove the 'rule of 9': an integer is divisible by 9 if and only if the sum of its integers is divisible by 9.
patriot [66]

Answer:

Rule of 9 : an integer is divisible by 9 if and only if the sum of its integers is divisible by 9.

Proof :

Let us consider a number x such that,

a=a_n......a_3a_2a_2

Where, a_0, a_1, a_2......,a_n are the the digits of the number x,

So, we can write,

x=a_0+a_1\times 10 + a_2\times 10^2 +a_3\times 10^3..........a_n\times 10^n

Let,

S=a_0+a_1+a_2+a_3+.......a_n

x-S=a_1(10-1)+a_2(10^2-1)+a_3(10^3-1)+.......a_n(10^n-1)

Since, a number is in the form of 10^k-1, where k is an positive integer, is always divisible by 9,

⇒ a_1(10-1), a_2(10^2-1), a_3(10^3-1),.......a_n(10^n-1) are divisible by 9.

⇒ x-S is divisible by 9,

⇒ If S is divisible by 9 ⇒ x must divisible by 9,

Or if x is divisible by 9 ⇒ S must divisible by 9.

Hence, proved...

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Step-by-step explanation:

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