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3 years ago

PLEASE ANSWER! URGENT

1 answer:

4 0

There would be 2034 students competing in 2012.

We can write a simple equation to find this answer. Let X, be the number of students starting in 2012. Then, we multiply by 0.9 and 1.1 to get to 2014. To work backwards, just divide by 1.1, then by 0.9.

2014 / 1.1 / 0.9 = 2034

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Barbara sells iced tea for $1. 49 per bottle and water for $1. 25 per bottle. She wrote an equation to find the number of bottle

Sedbober [7]

The correct option is The Barbara equation did not consider the number of bottles of water.

Given,

The selling price of iced tea is $1.49 per bottle.

The selling price of water is $1.25 per bottle.

Equation written by the Barbara,

$1.25x+1.49=100$

Here in this question Barbara multiply the selling price of bottle with the number of bottle using the coefficient x, But she did not use any coefficient for the bottle of water. The correct equation is,

$1.25x+1.49y=100$

where<em> y </em>is the number of water bottle sold to earn the profit of $100.

Hence the correct option is the Barbara equation did not consider the number of bottles of water.

For more about the linear equation, follow the link below-

brainly.com/question/11897796

3 0

3 years ago

Pls help zoom in if u can’t see

ale4655 [162]

Answer:

Step-by-step explanation:

because heat transfer from hot to cold but it does not transfer to something hotter or the same exact temp

5 0

3 years ago

Read 2 more answers

Calculați: [(8+8:8):9+(9+9:9):10-1]-1.

olchik [2.2K]

: sign means ratio
ration can said divide mark
<span>[(8+8:8):9+(9+9:9):10-1]-1
[(8+1):9+(9+1):10-1]-1
[(9:9)+(10:10)-1]-1
[1+1-1]-1
1-1
0</span>

3 0

3 years ago

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 of the following represents the graph and y-intercept of the function 3x +y= -2

laiz [17]

To find the Y intercept, you need to change this equation into y=mx + b format. You can do this by subtracting 3x from both sides of the equation. This results in the equation of

$y = 3x - 2$

If you are familiar with this format, it should be clear now that the Y intercept is - 2

8 0

3 years ago

Read 2 more answers