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3 years ago

13

Susan is buying supplies for a party. If spoons only come in bags of 8 and forks only come in bags of 6, what is the least numbe

r of spoons and the least number of forks she can buy so that she has the same number of each? asap and tell my how you got the answer please

1 answer:

igor_vitrenko [27]3 years ago

4 0

You can buy 6 bags of forks and 8 bags of spoons. All you simply have to do is find the common multiple of the two. You can just multiply them both if you can’t find any others or it’s easier

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Six more than twice a number is 24

Mars2501 [29]

Answer:

Step-by-step explanation:

I would say the answer is 6

6 + 6 than times by two is 24

3 0

2 years ago

Using the binomial theorem , obtain the expansion of :

andrezito [222]

Answer:

see explanation

Step-by-step explanation:

Expand both factors and collect like term

Using Pascal' triangle with n = 6 to obtain the coefficients

1 6 15 20 15 6 1

Decreasing powers of 1 from $1^{6}$ to $1^{0}$

Increasing powers of 3x from $(3x)^{0}$ to $(3x)^{6}$

$1+3x)^{6}$

= 1. $1^{6}$ $(3x)^{0}$ + 6. $1^{5}$ $(3x)^{1}$ + 15. $1^{4}$ $(3x)^{2}$ + 20. $1^{3}$ $(3x)^{3}$ + 15.1² $(3x)^{4}$ + 6. $1^{1}$ $(3x)^{5}$ + 1. $1^{0}$ $(3x)^{6}$

= 1 + 18x + 135x² + 540x³ + 1215 $x^{4}$ + 1458 $x^{5}$ + 729 $x^{6}$

--------------------------------------------------------------------------------------

$(1-3x)^{6}$

= 1. $1^{6}$ $(-3x)^{0}$ + 6. $1^{5}$ $(-3x)^{1}$ + 15. $1^{4}$ $(-3x)^{2}$ + 20. $1^{3}$ $(-3x)^{3}$ + 15.1² $(-3x)^{4}$ + 6. $1^{1}$ $(-3x)^{5}$ + 1. $1^{0}$ $(-3x)^{6}$

= 1 - 18x + 135x² - 540x³ + 1215 $x^{4}$ - 1458 $x^{5}$ + 729 $x^{6}$

----------------------------------------------------------------------------------

Collecting like terms from both expressions

$(1+3x)^{6}$ + $(1-3x)^{6}$

= 2 + 270x² + 2430 $x^{4}$ + 1458 $x^{6}$

----------------------------------------------------

(2)

Using Pascal's triangle with n = 5

1 5 10 10 5 1

Decreasing powers of 1 from $1^{5}$ to $1^{0}$

Increasing powers of 2x from $(2x)^{0}$ to $(2x)^{5}$

$(1+2x)^{5}$

= 1. $1^{5}$ $(2x)^{0}$ + 5. $1^{4}$ $(2x)^{1}$ + 10. $1^{3}$ $(2x)^{2}$ + 10. $1^{2}$ $(2x)^{3}$ + 5. $1^{1}$ $(2x)^{4}$ + 1. $1^{0}$ $(2x)^{5}$

= 1 + 10x + 40x² + 80x³ + 80 $x^{4}$ + 32 $x^{5}$

8 0

3 years ago

Solve the system of equations using row operations. Write your answer in the form (x, y).

Sonja [21]

Answer:

(5, - 4 )

Step-by-step explanation:

Given the 2 equations

2x + 3y = - 2 → (1)

3x - y = 19 → (2)

Multiplying (2) by 3 and adding to (1) will eliminate the y- term

9x - 3y = 57 → (3)

Add (1) and (3) term by term to eliminate y

(2x + 9x) + (3y - 3y) = (- 2 + 57), that is

11x = 55 ( divide both sides by 5 )

x = 5

Substitute x = 5 into either of the 2 equations and solve for y

Substituting into (1)

2(5) + 3y = - 2

10 + 3y = - 2 ( subtract 10 from both sides )

3y = - 12 ( divide both sides by 3 )

y = - 4

Solution is (5, - 4 )

8 0

3 years ago

55x ÷ 11x=?<br><br>solve for x

Liula [17]

Answer:

you cant solve for x but

55x/11x=5x

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

On sunday 1/16 of the people who visited for senior citizens, 3/16 or infants, 3/8 were children, and 3/8 for adults. How many o

Trava [24]

1/16=1,003, so

senior= 1003
infants=3009
children=6018
adults=6018

3 0

2 years ago

Read 2 more answers