What is the GCF of 44j^5k^4 and 121j^2k^6?

Mathematics

2 answers:

Leviafan [203]3 years ago

7 0

$44j^5k^4=2^2\cdot11\cdot j^5 \cdot k^4\\ 121j^2k^6=11^2\cdot j^2 \cdot k^6\\\\ \text{gcf}( 44j^5k^4 ,121j^2k^6)=11j^2k^4$

vampirchik [111]3 years ago

7 0

44j^5k^4=2^2\cdot11\cdot j^5 \cdot k^4\\ 121j^2k^6=11^2\cdot j^2 \cdot k^6\\\\ \text{gcf}( 44j^5k^4 ,121j^2k^6)=11j^2k^4

You might be interested in

What is twenty six million four hundred five thousand twenty in standard form?

andrew-mc [135]

Answer:

26,405,020

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Evaluate the variable expression when a=-4, b=2, c=-3, and d =4. b-3a/bc^2-d

gtnhenbr [62]

Answer:

Therefore, the variable expression when a=-4, b=2, c=-3, and d =4 is

$\dfrac{b-3a}{bc^{2}-d}=1$

Step-by-step explanation:

Evaluate:

$\dfrac{b-3a}{bc^{2}-d}$

When a=-4, b=2, c=-3, and d =4

Solution:

Substitute, a=-4, b=2, c=-3, and d =4 in above expression we get

$\dfrac{b-3a}{bc^{2}-d}=\dfrac{2-3(-4)}{2(-3)^{2}-4}\\\\=\dfrac{2+12}{18-4}\\\\$

$\dfrac{b-3a}{bc^{2}-d}=\dfrac{14}{14}=1$

Therefore, the variable expression when a=-4, b=2, c=-3, and d =4 is

$\dfrac{b-3a}{bc^{2}-d}=1$

6 0

3 years ago

Please, can someone help?

LekaFEV [45]

The area of that figure is 2453.25 square feet

First, you would need to find the area of the rectangle

(L)(W)
(70)(30)
2100

Then since it is only a semi circle, the formula is pi*r^2/2

You find the area of a full circle first

(3.14)(15^2)
(3.14)(225)
706.5

Then you divide that by 2 since it’s only half a circle

706.5/2 = 353.25

Finally you add that by the area of the rectangle

2100 + 353.25 = 2453.25 square feet

3 0

3 years ago

For the function F defined by F(x) = x2 – 2x + 4, find F(b+3).

Ivanshal [37]

Answer:

$\displaystyle F(b + 3) = b^2 + 4b + 7$