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patriot [66]
3 years ago
11

To which płace would you round to estimate 6,904-4,111 explain

Mathematics
1 answer:
Rainbow [258]3 years ago
8 0
To estimate you would round the numbers, so it would be 7,000 - 4,000 = 3,000. 
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25 POINTS!! Which of the following shows the number 10,000 written as a number having just three significant digits?
ivann1987 [24]

Answer:

B. 1.00 x 10 to the 4th power

Step-by-step explanation:

4 0
3 years ago
1 Point
alexandr1967 [171]

Answer:

Option C

Step-by-step explanation:

The given inequality is

5x-1 < 19 and -3-x+1\le1

Group similar terms to get:

5x < 19+1 and -x\le1-1+3

5x < 20 and -x\le3

Solve for x to get:

x < 4 and  x\ge -3

The correct choice is C

4 0
3 years ago
Read 2 more answers
Let N be the smallest positive integer whose sum of its digits is 2021. What is the sum of the digits of N + 2021?
kondor19780726 [428]

Answer:

10.

Step-by-step explanation:

See below for a proof of why all but the first digit of this N must be "9".

Taking that lemma as a fact, assume that there are x digits in N after the first digit, \text{A}:

N = \overline{\text{A} \, \underbrace{9 \cdots 9}_{\text{$x$ digits}}}, where x is a positive integer.

Sum of these digits:

\text{A} + 9\, x= 2021.

Since \text{A} is a digit, it must be an integer between 0 and 9. The only possible value that would ensure \text{A} + 9\, x= 2021 is \text{A} = 5 and x = 224.

Therefore:

N = \overline{5 \, \underbrace{9 \cdots 9}_{\text{$224$ digits}}}.

N + 1 = \overline{6 \, \underbrace{000 \cdots 000000}_{\text{$224$ digits}}}.

N + 2021 = 2020 + (N + 1) = \overline{6 \, \underbrace{000 \cdots 002020}_{\text{$224$ digits}}}.

Hence, the sum of the digits of (N + 2021) would be 6 + 2 + 2 = 10.

Lemma: all digits of this N other than the first digit must be "9".

Proof:

The question assumes that N\! is the smallest positive integer whose sum of digits is 2021. Assume by contradiction that the claim is not true, such that at least one of the non-leading digits of N is not "9".

For example: N = \overline{(\text{A})\cdots (\text{P})(\text{B}) \cdots (\text{C})}, where \text{A}, \text{P}, \text{B}, and \text{C} are digits. (It is easy to show that N contains at least 5 digits.) Assume that \text{B} \! is one of the non-leading non-"9" digits.

Either of the following must be true:

  • \text{P}, the digit in front of \text{B} is a "0", or
  • \text{P}, the digit in front of \text{B} is not a "0".

If \text{P}, the digit in front of \text{B}, is a "0", then let N^{\prime} be N with that "0\!" digit deleted: N^{\prime} :=\overline{(\text{A})\cdots (\text{B}) \cdots (\text{C})}.

The digits of N^{\prime} would still add up to 2021:

\begin{aligned}& \text{A} + \cdots + \text{B} + \cdots + \text{C} \\ &= \text{A} + \cdots + 0 + \text{B} + \cdots + \text{C} \\ &= \text{A} + \cdots + \text{P} + \text{B} + \cdots + \text{C} \\ &= 2021\end{aligned}.

However, with one fewer digit, N^{\prime} < N. This observation would contradict the assumption that N\! is the smallest positive integer whose digits add up to 2021\!.

On the other hand, if \text{P}, the digit in front of \text{B}, is not "0", then (\text{P} - 1) would still be a digit.

Since \text{B} is not the digit 9, (\text{B} + 1) would also be a digit.

let N^{\prime} be N with digit \text{P} replaced with (\text{P} - 1), and \text{B} replaced with (\text{B} + 1): N^{\prime} :=\overline{(\text{A})\cdots (\text{P}-1) \, (\text{B} + 1) \cdots (\text{C})}.

The digits of N^{\prime} would still add up to 2021:

\begin{aligned}& \text{A} + \cdots + (\text{P} - 1) + (\text{B} + 1) + \cdots + \text{C} \\ &= \text{A} + \cdots + \text{P} + \text{B} + \cdots + \text{C} \\ &= 2021\end{aligned}.

However, with a smaller digit in place of \text{P}, N^{\prime} < N. This observation would also contradict the assumption that N\! is the smallest positive integer whose digits add up to 2021\!.

Either way, there would be a contradiction. Hence, the claim is verified: all digits of this N other than the first digit must be "9".

Therefore, N would be in the form: N = \overline{\text{A} \, \underbrace{9 \cdots 9}_{\text{many digits}}}, where \text{A}, the leading digit, could also be 9.

6 0
3 years ago
Which graph represents this system?<br> 2x-5y =-5<br> y={x+1
Yuri [45]

Answer:

b

Step-by-step explanation:

4 0
3 years ago
Find each percent decrease of 80 to 64
Anastasy [175]
80-64=24
24 is 30% of 80 so a decrease of 30%
4 0
3 years ago
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