5 and 6 find the least common multiple of each set of number

Use the substitution method to solve the system of equations

Sergeeva-Olga [200]

Answer:

x=1/4

y = -2

(1/4, -2)

b

Step-by-step explanation:

4x -y =3

y =-8x

substitute y =-8x into the first equation. Everywhere you see y put 8x

4x -y =3

4x - (-8x) =3

4x+8x=3

combine like terms

12x = 3

divide by 12

12x/12x = 3 /12

x = 1/4

now we need to find y

y = -8x

y = -8 * 1/4

y = -2

8 0

3 years ago

Tiffany started her homework at 5:20, if it took him 45 minuets to get done, what time did he get finished

never [62]

Answer:

6:05

Step-by-step explanation:

5:20+40= 6:00

6:00+5= 6:05

5 0

3 years ago

Read 2 more answers

A soccer field for a youth league has an area of 4000 square yards and is 60% longer than it is wide what is the length of the f

olasank [31]

Let

width,w

length, l = 1.6w ...eqn 1

area, a = l × w ...eqn 2

subst for l in eqn 2

a = 1.6 w × w = 4000

4000 = 1.6 w^2

w^2 = 2500

w = 50

subst for w in eqn 1

l = 50 x 1.6 = 80

length = 80 yds

width = 50 yds

4 0

3 years ago

A deck of 52 cards contains 12 picture cards. If the 52 cards are distributed in a random manner among four players in such a wa

Mkey [24]

Answer:

The probability that each player will receive three picture cards = 0.0324

Step-by-step explanation:

As given,

A deck of 52 cards contains 12 picture cards

Remaining card = 52 - 12 = 40

So,

Total number of ways in which 12 picture card is distributed = $\frac{12!}{3! 3! 3! 3!}$

Now,

The Total number of ways in which Remaining cards are distributed = $\frac{40!}{10! 10! 10! 10!}$

So,

Total number of ways of getting 3 picture card and remaining card = $\frac{12!}{3! 3! 3! 3!}$ × $\frac{40!}{10! 10! 10! 10!}$

= $\frac{12! 40!}{(3!)^{4} (10!)^{4} }$

Now,

Total number of ways to distribute 52 cards so that each people get 13 card = $\frac{52!}{13! 13! 13! 13!} = \frac{52!}{ (13!)^{4} }$

∴ The probability = $\frac{\frac{12! 40!}{(3!)^{4} (10!)^{4} }}{\frac{52!}{(13!)^{4} }}$

= $\frac{12! 40!}{(3!)^{4} (10!)^{4} }$ × $\frac{(13!)^{4} }{ 52! }$

= $\frac{12! 40!}{(3!)^{4} (10!)^{4} }$ × $\frac{(13.12.11.10!)^{4} }{ 52.51.50.49.48.47.46.45.44.43.42.41.40! }$

= $\frac{12!}{(3!)^{4} }$ × $\frac{(13.12.11)^{4} }{ 52.51.50.49.48.47.46.45.44.43.42.41}$

= $\frac{479,001,600}{(6)^{4} }$ × $\frac{(1716)^{4} }{ 52.51.50.49.48.47.46.45.44.43.42.41}$

= 0.0324

∴ we get

The probability that each player will receive three picture cards = 0.0324

6 0

3 years ago

I’m stuck on this entire problem.

Hatshy [7]

A) No, because your bag has 8 marbles, and each time you take it out, you put it back in. So there is a chance that you can pick the same marble again, and not pick up a red marble that's already in the bag.

b) No, because you could still keep on picking up the same marble over and over again.

c) There is no absolutely certain number of times.

8 0

3 years ago