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3 years ago

Find the least common multiple of these two expressions. 4v^7 w^5 and 10v^2w^4y^8

Mathematics

1 answer:

8090 [49]3 years ago

7 0

Answer:

20v^7w^5y^8.

Step-by-step explanation:

The LCM of 4 and 10 is 20.

of v^7 and v^2 is v^7

of w^5 and w^4 is w^5

So the answer is 20v^7w^5y^8.

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What is the GCF of (8x-6) <br><br>sos T-T

makvit [3.9K]

Hi student, let me help you out! :)

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We are asked to find the G.C.F. of 8x-6.

$\triangle~\fbox{\bf{KEY:}}$

G.C.F. stands for Greatest Common Factor.

So what is the G.C.F. of 8x-6? We can list all of their common factors and then select the greatest one, like so:

$\longmapsto\bf{Factors\;of\;8x:}$ 1, 2, 4, 8, x

$\longmapsto\bf{Factors\;of\;-6:}$ -1, 1, 2, -3, 3, -6, 6

So the GCF is:

Now, we can also factor it out by placing it outside the parentheses ():

2(4x-3)

See, we factored it out by dividing both terms of the expression by the G.C.F.

$\ddot\bigstar$ Remember this...

$\fbox{\bf{We\;factor\;out\;the\;G.C.F.\;by\;dividing\;all\;the\;terms\;of\;the\;expression\;by\;it}}$

Hope this helps you out! :D

Ask in comments if any queries arise.

~Just a smiley person helping fellow students :)

6 0

1 year ago

Let omega be a complex number such that omega^3 = 1. Find all possible values of {1}{1 + \omega} + {1}{1 + \omega^2}. Enter all

Norma-Jean [14]

Answer:

1 is Answer.

Step-by-step explanation

$\frac{1}{1+omega^{2} } + \frac{1}{1+omega }\\$

= $\frac{(-1)*(omega + 1)}{omega^{2} }$

As we know that ω²+ω+1=0

Thus putting in above equation, we get

= $\frac{1}{(-1)*omega } + \frac{1}{(-1)*omega^{2} }$

Rearranging and simplifying:

= $\frac{-1}{omega } + \frac{-1}{omega^{2} }$

= $\frac{(-1)*(omega + 1)}{omega^{2} }$

= $\frac{(-1)*(- omega^{2} )}{omega^{2} }$

= 1 Answer

8 0

2 years ago

Pada hari kantin sebanyak 800 naskah kupon telah dijual,harga senaskah kupon masing masing rm 30 dan rm 50 .jumlah wang diperole

solniwko [45]

Answer:

Step-by-step explanation:

On the day of the canteen, 800 coupons were sold, the price of each coupon was RM 30 and RM 50 respectively. The amount of money earned from the sale of coupons was RM30000. How many copies of RM30 and RM50 coupons were sold?

Let:

RM 30 = x

RM 50 = y

x + y = 800 - - - (1)

30x + 50y = 30000 - - - (2)

From (1)

x = 800 - y

Put x = 800 - y in (2)

30(800 - y) + 50y = 30000

24000 - 30y + 50y = 30000

24000 + 20y = 30000

20y = 30000 - 24000

20y = 6000

y =

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2 years ago

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Answer:

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2 years ago

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