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Dmitry_Shevchenko [17]

3 years ago

8

Graph y minus 3 equal one half (x+2)

1 answer:

Scorpion4ik [409]3 years ago

6 0

This is the graph for y-3=1/2(x+2)

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Anestetic [448]

Find multiples of 30 and 40:

30: 30, 60, 90, 120, 150

40: 40, 80, 120, 160

The least common multiple is 120, so every 120 minutes they willl both ring.

1 hour = 60 minutes:

120/60 = 2 hours.

The bells ring together every 2 hours.

8 + 2 = 10 am

10 + = 12pm

12 + 2 = 2 pm

The bells will ring together at am, 12 pm and 2 pm.

6 0

3 years ago

Read 2 more answers

Raina must choose a number between 55 and 101 that is a multiple of 2, 8 and 10. Write all the numbers that she could choose. If

Aleonysh [2.5K]

Answer:

The only possible number is $80$ .

Step-by-step explanation:

The number in question needs to be a multiple of all three of $2$ , $8$ , and $10$ . As a result, it must also be a multiple of the least common multiplier (lcm) of the three number.

Start by finding the least common multiplier of the three numbers.

Factor each number into its prime components:

$2$ is a prime number itself.
$8 = 2^3$ .
$10 = 2 \times 5$ .

The only prime factors are $2$ and $5$ .

The greatest power of $2$ among the three numbers is $3$ .
The greatest power of $5$ among the three numbers is $1$ .

Therefore, the least common multiplier of the three number should be the product of $2^3$ and $5$ . That's equal to $2^3 \times 5 = 8 \times 5 = 40$ .

In other words, the number (or numbers) in question could be written in the form $40\, k$ , where $k$ is an integer.

The question requires that this number be between $55$ and $101$ . In other words,

$55 \le 40\, k \le 101$ .

The goal is to find the possible values of $k$ . Note that from integer division by $40$ ,

$55 = 1 \times 40 + 15$ , and
$101 = 2 \times 40 + 21$ .

The inequality becomes:

$1 \times 40 + 15 \le 40\, k \le 2 \times 40 + 21$ .

However,

$1 \times 40 < 55 = 1 \times 40 + 15$ , and
$2 \times 40 + 21 = 101 < 3 \times 40$ .

Hence,

$1 \times 40 < 1\times 40 + 15 \le 40\, k \le 2 \times 40 + 21 < 3 \times 40$ .

$1 \times 40 < 40\, k < 3 \times 40$ .

Divide by the positive number $40$ to obtain:

$1 < k < 3$ .

Since $k$ is an integers, $k = 2$ .

Indeed, $40 \, k = 80$ is between $55$ and $101$ .

Therefore, $80$ is the number in question.

5 0

4 years ago

Read 2 more answers

Find the sum of first 12 terms of the series 5,9,13,17 ?

Lubov Fominskaja [6]

27 I think ou this is weird what exactly is the subject

6 0

3 years ago

Denise has a roll of ribbon that is 9 feet

zvonat [6]

Your answer would be -12 because of her answers

7 0

3 years ago

What is the distance between the points (1, -4) and (1, 0)

tino4ka555 [31]

For this case we have that the distance between two points is given by:

$d = \sqrt{(y2-y1)^2+(x2-x1)^2}$

Given the points:

$(x1, y1) = (1, -4)(x2, y2) = (1,0)$

We substitute in the formula:

$d = \sqrt{(0 - (- 4)) ^ 2 + (1-1) ^ 2}$

$d = \sqrt{(4 ^ 2 + 0 ^ 2)}$

$d=\sqrt{16}$

$d = 4$

So, the distance is $d = 4$

Answer:

$d = 4$

5 0

3 years ago