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White raven [17]
2 years ago
13

X + 1 = x + 1 solve for x plz help.

Mathematics
2 answers:
Klio2033 [76]2 years ago
8 0

Answer:

Infinite amount of solutions.

Step-by-step explanation (check image for written down steps, excuse bad handwriting):

Well, in order to solve this, you would subtract x on both sides, leaving you with 1 = 1

Does 1 = 1? Yes, it does, so there is an infinite amount of solutions. ∞

Notes:

Check image

Yanka [14]2 years ago
7 0

Answer:

2 ??????

Step-by-step explanation:

sorry if it is wrong

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A week except Sunday = 6 days

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How many pairs of consecutive natural numbers have a product of less than 40000? I am in 5th grade. This is supposed to be easy
Ugo [173]

Answer:

There are 199 pairs of consecutive natural numbers whose product is less than 40000.

Step-by-step explanation:

We notice that such statement can be translated into this inequation:

n \cdot (n+1) < 40000

Now we solve this inequation to the highest value of n that satisfy the inequation:

n^{2}+n < 40000

n^{2}+n -40000

The Quadratic Formula shows that roots are:

n_{1,2} = \frac{-1\pm\sqrt{1^{2}-4\cdot (1)\cdot (-40000)}}{2\cdot (1)}

n_{1,2} = -\frac{1}{2}\pm \frac{1}{2} \cdot \sqrt{160001}

n_{1} = -\frac{1}{2}+\frac{1}{2}\cdot \sqrt{160001}

n_{1} \approx 199.501

n_{2} = -\frac{1}{2}-\frac{1}{2}\cdot \sqrt{160001}

n_{2} \approx -200.501

Only the first root is valid source to determine the highest possible value of n, which is n_{max} = 199. Each natural number represents an element itself and each pair represents an element as a function of the lowest consecutive natural number. Hence, there are 199 pairs of consecutive natural numbers whose product is less than 40000.

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2 years ago
Need it ASAP the assignment is due tonight help pls
Gre4nikov [31]
-5/2x + b = y
Check photo for explanation :)
————————-

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