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3 years ago

Find the Least Common Multiple of 15w and 6w^2.

2 answers:

7 0

For this case, we have that by definition, the LCM of two or more natural numbers, is the smallest natural number that is a common multiple of all of them.

Then the LCM of 15 and 6 is the smallest positive integer that divides the numbers 15 and 6 without leaving a residue.

Multiples of each number:

15: 15,30,45 ...

6: 6,12,18,24,30

So, the LCM is $30w ^ 2$

Answer:

$30w ^ 2$

Lina20 [59]3 years ago

3 0

15= 1,3,5,15

6^2= 36

36=1,2,3,4,6....

LCM of 15W and 6W^2=3

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Please answer this and thank you<br><br> If DE=4x+10, EF=2-1, and DF=9x-15, find DF.

professor190 [17]

I hope that 2-1 means 21.

DE + EF = DF
4x + 10 + 21 = 9x - 15
4x + 31 = 9x - 15
31 = 9x - 4x - 15
31 = 5x - 15
46 = 5x
x = 9 1/5 = 9.2

You want 9x - 15
9*9.2 - 15
82.8 - 15
67.8 <<<<< answer.
If I have given the wrong meaning to 2-1 (which could mean 1) leave another note and I will correct my work.

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3 years ago

Mrs. Nelson has a rectangular flower box that is 5 feet long and 2 feet tall. She wants the width of the box to no more than 5 f

d1i1m1o1n [39]

Volume = length x width x height.
if the width has to be a whole number no more than 5 then the only options for the width is 1, 2, 3, 4, or 5.
V=5x1x2
V=10
V=5x2x2
V=20
V=5x3x2
V=30
V=5x4x2
V=40
V=5x5x2
V=50
the possible volumes are as follows: 10, 20, 30, 40, and 50.
make sure in your answer that you label your volumes so you write ft^3 after every volume answer.

7 0

3 years ago

Peggy is thinking of a number such that when twice the number is added to three times one more than the number she gets the same

mars1129 [50]

Answer:

-7

Step-by-step explanation:

Lets assume the number Peggy is thinking of be "x".

Now as given, when twice the number is added to three times one more than the number. We can write it as

∴ $2x+3(x+1)$ --- Equation 1

Again, it is given that Peggy gets the same result as when she multiplies four times one less than the number.

∴ $4(x-1)$ -- equation 2

Next, we can equate both the equation 1 and 2 as it is given that result is same.

$2x+3(x+1)= 4(x-1)$

Let´s distribute 3 into $(x+1)$ and 4 into $(x-1)$

⇒ $2x+3x+3=4x-4$

⇒ $5x+3=4x-4$

Subtract both side by 4x and 3

∴ $x=-7$

∴ Peggy was thinking of -7

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ahrayia [7]

Answer:

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