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deff fn [24]
4 years ago
7

The value of the digit in the hundreds placed in the number 653841 is 1/10 the value of the digit in the thousand place in which

number?

Mathematics
1 answer:
kondor19780726 [428]4 years ago
6 0
The complete question in the attached figure

we have that
the value of the digit in the hundreds place in the number 653,841 is---> 800

case  A) 748,917
the value of the digit in the thousands place------> 8000
1/10 the value of the digit is-------> (1/10)*8000=800
800=800---------------> is ok

case  B) 749,817
the value of the digit in the thousands place------> 9000
1/10 the value of the digit is-------> (1/10)*9000=900
900 is not 800

case  C) 784,817
the value of the digit in the thousands place------> 4000
1/10 the value of the digit is-------> (1/10)*4000=400
400 is not 800

case  D) 797,481
the value of the digit in the thousands place------> 7000
1/10 the value of the digit is-------> (1/10)*7000=700
700 is not 800


the answer is
the option A) 748,917

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Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.
Verdich [7]

Answer:

A) -18, 3, 24, 45, ...   f(1) = -18, f(n) = f(n-1) + 21 for n = 2,3,4,5,...

B) 11, 33, 55, 77, ...  f(1) = 11, f(n) = f(n-1) + 22 for n>1

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Step-by-step explanation:

-18, 3, 24, 45, ...

Let's take look at the possible difference patterns between terms:

-18 -> 3 ==> +21 or /-6

3 -> 24 ==> +21 or *8

We see we have a common pattern possible... +21

Which is confirmed from 24 to 45.

So the function will be f(1) = -18 of course

f(n) = f(n-1) + 21 for n = 2,3,4,5,...

11, 33, 55, 77, ...

Let's take a look at the possible difference patterns here again:

11 --> 33 ==> +22 or *3

33 --> 55 ==> + 22 or *5/3

Again, pattern of +22 confirmed with 55 -> 77

f(1) = 11

f(n) = f(n-1) + 22 for n>1

-18, -108, -648, -3,888, ...

Possible patterns:

-18 --> -108 ==> -90 or *6

-108 --> -648 ==> -540 or *6

Pattern of *6 confirmed with -648 and -3,888.

f(1) = -18

f(n) = f(n-1) * 6 for n > 1

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Answer:

See below

Step-by-step explanation:

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