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astraxan [27]
3 years ago
7

I have 30 ones, 82 thousands, 4 hundred thousands, 60 tens, and 100 hundreds. What number am I?

Mathematics
1 answer:
stich3 [128]3 years ago
7 0
This problem tackles the place values of numbers. From the rightmost end of the number to the leftmost side, these place values are ones, tens, hundreds, thousands, ten thousands, hundred thousands, millions, ten millions, one hundred millions, and so on and so forth. My idea for the solution of this problem is to add up all like multiples. In this problem, there are 5 multiples expressed in ones, thousands, hundred thousands, tens and hundreds. Hence, you will add up 5 like terms. The solution is as follows

30(1) + 82(1,000) + 4(100,000) + 60(10) + 100(100)

The total answer is 492,630. Therefore, the number's identity is 492,630.
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H(x) = x4+ 2x3- 10x2 -18x+9
vichka [17]
Let's group the cubic function: h(x)=x^4+2x^3-10x^2-18x+9=(x-3)(x+3)(x^2+2x-1).

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x^2+2x-1=0
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The roots of the cubic function are \pm3, -1-\sqrt{2} and \sqrt{2}-1.
5 0
3 years ago
Read 2 more answers
Find the missing side​
Yuri [45]

Answer:

Step-by-step explanation:

10

the answer is 10

7 0
3 years ago
What is the result when we simplify the expression
Ostrovityanka [42]

Answer:

Step-by-step explanation:

Here you go mate

Step 1

(1+1x)(1-2x+1)(1+2x-1)  Equation/Question

Step 2

(1+1x)(1-2x+1)(1+2x-1)  Remove parenthesis

-2x^2+2-4x^3+4x+2x^2-2

Step 3

-2x^2+2-4x^3+4x+2x^2-2  Simplify

-4x^3+4x

Answer

-4x^3+4x

Hope this helps

7 0
3 years ago
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