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Semenov [28]
2 years ago
7

Choose the most convenient method to graph the line 2x+5y=−10.

Mathematics
1 answer:
Serjik [45]2 years ago
6 0

Answer:

graph the line using the slope and y-intercept or two points

slope:-2/5

y-intercept:(0,-2)

x | y

-5 | 0

0 | -2

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The least common multiple of two numbers is 60. the prime factorization of one number is 3 times 5. what is the prime factorizat
gulaghasi [49]
The first number would be 3×5=15. 
Another number could have different values. it could be
4=2×2
12=2×2×3
20=2×2×5
60=2×2×3×5
7 0
3 years ago
What multiple of 7 is also a factor of 7
enyata [817]
<span>The answer is 7. Seven is a prime number, which means that its multiples are 1 and itself (7): 7 = 1 * 7. To find out which one is a factor of 7, we should divide each of them by 7. 1/7 will be a decimal number and 7/7 = 1, which is a whole number. Thus, 7 will be a multiple of 7 and also a factor of 7.Hope this helps. Let me know if you need additional help!</span>
7 0
3 years ago
Triangle ABC has coordinates A (0, 1) B (0, 2) and C (3,2). If Triangle ABC is equivalent to triangle EDF, what is the measure o
Elanso [62]

Answer:

  3

Step-by-step explanation:

Segment BC corresponds to segment DF. The length of BC is the distance between coordinates (0, 2) and (3, 2). These points are on the same horizontal line (y=2), so the distance between them is the difference of their x-coordinates: 3 - 0 = 3.

7 0
3 years ago
Read 2 more answers
801,504 round to the nearest thousand​
Gelneren [198K]

Answer:

802,000

Step-by-step explanation:

The number in the thousands place is 1, and the number to the right of it is 5. And if the number to the right is greater than or equal to 5 the place value that you are rounding goes up by one.

3 0
3 years ago
A school has 1800 students and 1800 light bulbs, each with a pull cord and all in a row. All the lights start out off. The first
Gnom [1K]

<u>Solution-</u>

A school has 1800 students and 1800 light bulbs, each with a pull cord and all in a row.

As all the lights start out off, in the first pass all bulbs will be turned on.

In the second pass all the multiples of 2 will be off and rest will be turned on.

In the third pass all the multiples of 3 will be off, but the common multiple of 2 and 3 will be on along with the rest. i.e all the multiples of 6 will be turned on along with the rest.

In the fourth pass 4th light bulb will be turned on and so does all the multiples of 4.

But, in the sixth pass the 6th light bulb will be turned off as it was on after the third pass.

This pattern can observed that when a number has odd number of factors then only it can stay on till the last pass.

1 = 1

2 = 1, 2

3 = 1, 3

<u>4 = 1, 2, 4</u>

5 = 1, 5

6 = 1, 2, 3, 6

7 = 1, 7

8 = 1, 2, 4, 8

9 = 1, 3, 9

10 = 1, 2, 5, 10

11 = 1, 11

12 = 1, 2, 3, 4, 6, 12

13 = 1, 13

14 = 1, 2, 7, 14

15 = 1, 3, 5, 15

16 = 1, 2, 4, 8, 16

so on.....

The numbers who have odd number of factors are the perfect squares.

So calculating the number of perfect squares upto 1800 will give the number of light bulbs that will stay on.

As,  \sqrt{1800} =42.42  , so 42 perfect squared numbers are there which are less than 1800.

∴ 42 light bulbs will end up in the on position. And there position is given in the attached table.

7 0
3 years ago
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