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2 years ago

893 hundreds x 85 tens =

Mathematics

2 answers:

cricket20 [7]2 years ago

8 0

The answer is 75,905

Sveta_85 [38]2 years ago

6 0

Answer:

75,905,000 or 75,905 thousands.

Step-by-step explanation:

We are asked to find the product of the given quantities.

893 hundreds would be $893\times 100=89300$ .

85 tens would be $85\times 10=850$ .

The product of both numbers would be:

$89,300\times 850=75,905,000$ .

Therefore, our required product would be 75,905,000 or 75,905 thousands.

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Step-by-step explanation:

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WX=XY-----(1)

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A car uses 3t gallons of gasoline to travel 5 miles.

Nadusha1986 [10]

Answer: a) $\frac{24}{5}t gallons$ are needed to travel 8 miles

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2 years ago

The manager of a symphony in a large city wants to investigate music preferences for adults and students in the city. Let pA rep

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6 0

1 year ago

The first term of a finite geometric series is 6 and the last term is 4374. The sum of all the term os 6558. find the common rat

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A geometric series is written as $ar^n$ , where $a$ is the first term of the series and $r$ is the common ratio.

In other words, to compute the next term in the series you have to multiply the previous one by $r$ .

Since we know that the first time is 6 (but we don't know the common ratio), the first terms are

$6, 6r, 6r^2, 6r^3, 6r^4, 6r^5, \ldots$ .

Let's use the other information, since the last term is $4374 > 6$ , we know that $r>1$ , otherwise the terms would be bigger and bigger.

The information about the sum tells us that

$\displaystyle \sum_{i=0}^n 6r^i = 6\sum_{i=0}^n r^i = 6558$

We have a formula to compute the sum of the powers of a certain variable, namely

$\displaystyle \sum_{i=0}^n r^i = \cfrac{r^{n+1}-1}{r-1}$

So, the equation becomes

$6\cfrac{r^{n+1}-1}{r-1} = 6558$

The only integer solution to this expression is $n=6, r=3$ .

If you want to check the result, we have

$6+6*3+6*3^2+6*3^3+6*3^4+6*3^5+6*3^6 = 6558$

and the last term is

$6*3^6 = 4374$

7 0

2 years ago