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elixir [45]
3 years ago
9

Find the length of a rectangular lot that has a perimeter of 594 inches if the length is 3 inches lass than twice the width

Mathematics
2 answers:
Ganezh [65]3 years ago
7 0
A) 2L + 2W = 594
B) L = 2W +3

B) L -2W = 3 Then adding to A
A) 2L + 2W = 594
3L = 597
L = 199

2W = L+3
W = (L + 3)/2
W = 100

Double Check
2W + 2L = 594 Perimeter

igomit [66]3 years ago
3 0
Length equals 29 that's the answer
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Rewrite the expression g+1/6d-3/4g+1/9g-g+1/4d in standard from by combining like terms I need this ASAP AND I NEED ALL STEPS TO
GarryVolchara [31]

Answer:

(5/12)d - (23/36)g

Step-by-step explanation:

First you can eliminate g and -g to get (1/6)d - (3/4)g + (1/9)g + (1/4)d. Then you need to get common denominators to add like terms together.

1/6 = 4/24 and 1/4 = 6/24. Add them together to get (10/24)d or (5/12)d.

-3/4 = -27/36 and 1/9 = 4/36. Add them together to get (-23/36)g.

So in standard form, (5/12)d - (23/36)g

5 0
3 years ago
Plz help will mark brainliest :)
shepuryov [24]

Answer:

an odd    an even

Step-by-step explanation:

plz mark as brainliest. my first answer

3 0
3 years ago
Show work 29 points pls help
Tanzania [10]

Answer:

a

Step-by-step explanation:

The equation of a proportional relationship is

y = kx ← k is the constant of proportionality

Divide both sides by x

\frac{y}{x} = k

To find k use any ordered pair from the table

Using (2, \frac{5}{2} ) , then

k = \frac{2}{\frac{5}{2} } = 2 × \frac{2}{5} = \frac{4}{5}

7 0
3 years ago
What are the prime factors for 25
Anni [7]
1,5, and 25. 1*25=25 and 5*5=25
4 0
3 years ago
Read 2 more answers
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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