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madreJ [45]
3 years ago
11

A^2 + 3b ÷ c - 2d A = 3 , B = 8 , C = 2, D = 5 please help me!

Mathematics
1 answer:
s2008m [1.1K]3 years ago
6 0

Answer:

11

Step-by-step explanation:

Substitute the given values into the expression

3² + 3(8) ÷ 2 - 2(5)

= 9 + 24 ÷ 2 - 10 ← evaluate division before addition/ subtraction

= 9 + 12 - 10

= 21 - 10

= 11

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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
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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}

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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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