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tensa zangetsu [6.8K]
3 years ago
9

There are some children playing at the park when 4 more arrive and begin playing. Now there are 28 children playing at the park.

How many children were originally playing at the park?
Mathematics
1 answer:
lisov135 [29]3 years ago
5 0
Currently there are 28 children playing. Until a moment ago, 4 of those 28 werent there. To find out how many there were a moment ago, we have to exclude those 4 kids that just came — or subtract the number of kids that just came from the number of kids currently playing.

28-4=24

We can check: 24+4=28
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Help answer please!!
12345 [234]

Answer:

h = 19.7 ft.

Step-by-step explanation:

Use the Pythagoras Theorem

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Now we take square root on both sides,

\sqrt{387.75} =\sqrt{h^2}\\19.69=h\\h=19.7\ ft\\

so the value of h equals 19.7 ft.

6 0
3 years ago
PLEASE SHOW ALL THE STEPS THAT YOU USE TO SOLVE THIS PROBLEM
Mademuasel [1]

Answer:

{x = 1 , y=1, z=0

Step-by-step explanation:

Solve the following system:

{-2 x + 2 y + 3 z = 0 | (equation 1)

{-2 x - y + z = -3 | (equation 2)

{2 x + 3 y + 3 z = 5 | (equation 3)

Subtract equation 1 from equation 2:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x - 3 y - 2 z = -3 | (equation 2)

{2 x + 3 y + 3 z = 5 | (equation 3)

Multiply equation 2 by -1:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+3 y + 2 z = 3 | (equation 2)

{2 x + 3 y + 3 z = 5 | (equation 3)

Add equation 1 to equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+3 y + 2 z = 3 | (equation 2)

{0 x+5 y + 6 z = 5 | (equation 3)

Swap equation 2 with equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y + 6 z = 5 | (equation 2)

{0 x+3 y + 2 z = 3 | (equation 3)

Subtract 3/5 × (equation 2) from equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y + 6 z = 5 | (equation 2)

{0 x+0 y - (8 z)/5 = 0 | (equation 3)

Multiply equation 3 by 5/8:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y + 6 z = 5 | (equation 2)

{0 x+0 y - z = 0 | (equation 3)

Multiply equation 3 by -1:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y + 6 z = 5 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Subtract 6 × (equation 3) from equation 2:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y+0 z = 5 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Divide equation 2 by 5:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+y+0 z = 1 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Subtract 2 × (equation 2) from equation 1:

{-(2 x) + 0 y+3 z = -2 | (equation 1)

{0 x+y+0 z = 1 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Subtract 3 × (equation 3) from equation 1:

{-(2 x)+0 y+0 z = -2 | (equation 1)

{0 x+y+0 z = 1 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Divide equation 1 by -2:

{x+0 y+0 z = 1 | (equation 1)

{0 x+y+0 z = 1 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Collect results:

Answer:  {x = 1 , y=1, z=0

6 0
3 years ago
How many ways can eight letters be arranged into groups of five where order matters and the first two letters are already chosen
fenix001 [56]

Answer:

120

Step-by-step explanation:

Since we're dealing with a problem where the order matters and the first two letters are already chosen we need to subtract the number of letters and the number of available slots per group.

We use the permutation formula to find the answer, but before that let's check values.

n = 8

k = 5

Now since there are two letters already chosen we have to deduct two from both the value of n and k.

n = 6

k = 3

Now we can use the permutation formula:

_{n}P_{k}=\dfrac{n!}{(n-k)!}

_{6}P_{3}=\dfrac{6!}{6-3)!}

_{6}P_{3}=\dfrac{6!}{3!}

_{6}P_{3}=\dfrac{6*5*4*3*2*1}{3*2*1}

The 3*2*1 cancels out and leaves us with:

_{6}P_{3}=6*5*4

_{6}P_{3}=120

So there are 120 possible ways to arrange eight letters into groups of five where order matters and the first two letters are already chosen.

6 0
4 years ago
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In a group discussion, which option is most clearly a clarifying question?
Mariulka [41]

Answer:

your answer should be d

Step-by-step explanation:

5 0
4 years ago
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UkoKoshka [18]

Answer:

68

Step-by-step explanation:

Let x=adult tickets and y=child tickets

1) 15x+11y=1889

2) x+y=147

multiply equation 1 by -1/11 on both sides to get:

-(15/11)x-y=-171.73

add new equation with equation 2 to get

-4/11x=-24.73

Isolate x to get:

x=67.99999

Round up to get 68 adult tickets

6 0
3 years ago
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