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DerKrebs [107]
3 years ago
5

(1 point) How many ways can you give 9 (identical) apples to your 5 favourite Mathematics lecturers (without any restrictions)?

Mathematics
1 answer:
cricket20 [7]3 years ago
7 0

Answer:

126

Step-by-step explanation:

Total number of apples = 9

Number of favourite Mathematics lecturers = 5

To find number of ways in which 9 (identical) apples can be given to your 5 favourite Mathematics lecturers, find C(9,5)

Combination refer to selection of items such that order does not matter.

Use C(n,p)=\frac{n!}{p!(n-p)!}

Put n=9,p=5

C(9,5)=\frac{9!}{5!(9-5)!}\\=\frac{9!}{5!4!}\\=\frac{9(8)(7)(6)5!}{5!4!} \\=\frac{9(8)(7)(6)}{4(3)(2)(1)}\\=126

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1. (2,7); m=-4<br> What’s the answer?
Nataly_w [17]

Answer:

4x + y = 15 or y = -4x + 15

Step-by-step explanation:

7 = 2[-4] + b

-8

15 = b

y = -4x + 15

If you want it in <em>Standard </em><em>Form</em>:

y = -4x + 15

+4x +4x

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4x + y = 15 >> Line in <em>Standard Form</em>

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4 0
3 years ago
JIMMMMMM
Vitek1552 [10]
>= means greater than or equal to
<= means less than or equal to

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

Part A

The graph of y >= -3x+3 will have a solid boundary line and the shading will be above the boundary line.

The boundary line y = -3x+3 has a negative slope so it moves down as you read it from left to right. It goes through the points (0,3) and (1,0)

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

The graph of y < (3/2)x - 6 will have a dashed or dotted boundary line. The shading is below the boundary.

The graph y = (3/2)x-6 goes through the two points (0,-6) and (2,-3)

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

If you graph both y >= -3x+3 and y < (3/2)x - 6 together, you get what you see in the attached image. This solution shaded region is the result of the overlapping prior shaded regions. 

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

Part B

Plug (x,y) = (-6,3) into each inequality to see if we get a true inequality or not

For the first inequality we have
y >= -3x+3
3 >= -3*(-6)+3
3 >= 18+3
3 >= 21
which is false. The value 3 is not larger or equal to 21. So right off the bat we know that (-6,3) is NOT a solution. It is NOT in the solution region.

Let's check the other inequality just for the sake of completeness
y < (3/2)x - 6
3 < (3/2)*(-6) - 6
3 < -9 - 6
3 < -15
this is also false. The value -15 is smaller than 3, since it is to the left of 3

We're given more evidence that (-6,3) is NOT in the solution area. It is outside of both shaded areas. 

7 0
3 years ago
Read 2 more answers
Help plz i give brainliest
Natasha_Volkova [10]

Answer: Hope this helps :)

\frac{1}{5} b^{27}

<h2><u><em>Mark me as Brainliest please</em></u></h2>
6 0
2 years ago
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Can the formula F=ma be a direct variation equation? If so, which variable would have to be held constant?
ANEK [815]
Yes it can be a direct variation.  it follows the form y = kx
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6 0
3 years ago
Which ordered pair is a solution to the system of linear equations One-half x minus three-fourths y = StartFraction 11 Over 60 E
s344n2d4d5 [400]

Answer:

\left(\dfrac{2}{3},\dfrac{1}{5}\right)

Step-by-step explanation:

Given the system of two equations:

\left\{\begin{array}{l}\dfrac{1}{2}x-\dfrac{3}{4}y=\dfrac{11}{60}\\ \\\dfrac{2}{5}x+\dfrac{1}{6}y=\dfrac{3}{10}\end{array}\right.

Multiply the first equation and the second equation by 60 to get rid of fractions:

\left\{\begin{array}{l}30x-45y=11\\ \\24x+10y=18\end{array}\right.

Now multiply the first equation by 4 and the second equation by 5:

\left\{\begin{array}{l}120x-180y=44\\ \\120x+50y=90\end{array}\right.

Subtract them:

(120x-180y)-(120x+50y)=44-90\\ \\120x-180y-120x-50y=-46\\ \\-230y=-46\\ \\y=\dfrac{46}{230}=\dfrac{1}{5}

Substitute it into the first equation:

30x-45\cdot \dfrac{1}{5}=11\\ \\30x-9=11\\ \\30x=11+9\\ \\30x=20\\ \\x=\dfrac{2}{3}

The solution is \left(\dfrac{2}{3},\dfrac{1}{5}\right)

4 0
3 years ago
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