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3 years ago

Just a bunch of easy ones

Mathematics

2 answers:

Scilla [17]3 years ago

5 0

Answer:

1. 5lb bag is cheaper

2. 6 mpm (minutes per miles)

3. Brycen

4. 18 mpm faster

5. 32 total shapes

6. 75

7. not enough info

8. 50 miles

9. 2

10. 2

Step-by-step explanation:

3lb=$3.54=1lb=$1.18

5lb=$5.85=1lb=$0.85

12 minutes/2 miles=6 mpm

24 minutes/1 mile=24 mpm

24-6=18 mpm faster

25% of 100= 4

4×8=32

60% of 80= 75

process of elimination:

24.5 is less than 35 so no

50×.7=35, total miles is 50

d1i1m1o1n [39]3 years ago

5 0

For Lin if she runs 3 laps in 6 minutes. you need to divided 3/6=2

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Solve for r.<br> -1r + 1 > 7

Alexeev081 [22]

Answer:

r < − 6

Step-by-step explanation:

7 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 $9$ .

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3 years ago

Which equation goes with the graph

Rus_ich [418]

Link is scam. literally cap. don’t use it or what do ever!

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3 years ago

What is the solution to the equation 2(m - 8) = 4(m + 6) ? m =

ExtremeBDS [4]

The first thing that we want to do is simplify both side of this equation.

2(m-8).
The first thing we have to do, is multiple 2 by m and -8, also called distributing. So, when we do that, we get
2(m-8)=2*m + 2*-8 = 2m-16

4(m+6)
Now, we do the same thing we did for the first parenthesis, and that's multiple 4 by m and 6.
4(m+6)=4*m + 4*6 = 4m+24

Now, we need to get m by itself on one side. So, lets bring the equation from the right to the left

2m-16 = 4m + 24 (bring 4m to the other side by subtracting)
2m-16-4m = 24 (bring -16 to the right by adding)
-2m=24+16
-2m=40 (divide both sides by -2 to get your value for m)
m=40/-2 = -20

So, our answer is m = -20

3 0

3 years ago

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Tania has 6 tiles with pictures of plants and 3 tiles with pictures of animals. Tania keeps all the tiles on a mat with the pict

Natalka [10]

Initially there were 6 tiles with pictures of plants and 3 tiles with pictures of animals.
She removes one tile with the picture of a plant, so now there are 5 tiles with pictures of plants and 3 tiles with pictures of animals. There are 5+3=8 tiles in total.

The probability that the second tile has a plant on it is the number of tiles with pictures of plants (5) divided by the total number of tiles (8).

The probability is $\frac{5}{8}$ .

5 0

3 years ago