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mamaluj [8]
3 years ago
5

Please answer this question now

Mathematics
2 answers:
Kipish [7]3 years ago
6 0

Answer:

u ≈ 13.7

Step-by-step explanation:

Using the Cosine rule in Δ STU, that is

u² = s² + t² - 2stcosU

Here s = 21, t = 9 and U = 28°, thus

u² = 21² + 9² - (2 × 21 × 9 × cos28°)

   = 441 + 81 - 378 cos28°

   = 522 - 378 cos28° ( take the square root of both sides )

u = \sqrt{522-378cos28}

   ≈ 13.7 ( to the nearest tenth )

yarga [219]3 years ago
3 0

Answer:

\boxed{u = 13.7}

Step-by-step explanation:

Using cosine rule

c^2 = a^2+b^2-2ab\ CosC

Here c = u, a = 9 , b = 21 and C = 28

u^2 = 9^2+21^2-2(9)(21)\ Cos 28\\u^2 = 81+441-(378)(0.88)\\u^2 = 522 - 333.75\\u^2 = 188.24

Taking sqrt on both sides

u = 13.7

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Answer: -14

Step-by-step explanation: Here we're asked to simplify -|-14| so we

start by simplifying the absolute value of -14.

Remember that the absolute value of

any positive or negative integer is positive.

So the absolute value of -14 is 14.

So we write (14) in parenthses like I have just done.

Next, we bring down the negative sign

that was outside the absolute value.

So we're left with -(14) or -14.

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4 years ago
Hich statement about the student's answer is correct?
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When the Dragons have lost their most recent game, 200 fans buy tickets. For each consecutive win the
Andrei [34K]

Answer:

A function that gives the of tickets is given by

t(w)=200(1.1)^u

where u is number of consecutive wins

Step-by-step explanation:

Given that,

Initial amount of fans is =200.

For each wins, the number ticket sells is also increased by a factor of 1.1.

After 1^{st} win, the number of tickets fans will be 200×1.1

After 2^{nd} win, the number of tickets fans will be 200×1.1×1.1

                                                                              =200(1.1)²

After 3^{rd} win, the number of tickets fans will be 200×1.1×1.1×1.1

                                                                             =200(1.1)³

and so on.

After u^{th} win, the number of tickets fans will be 200(1.1)^u.

A function that gives the of tickets is given by

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8 0
4 years ago
Read 2 more answers
You toss three 6-sided dice and record the sum of the three faces facing up. a) Describe precisely a sample space S for this exp
timama [110]

Answer:

a.)Sample space means all possible outcomes, since we know the dice are 6 face, then the sample space becomes all possible outcomes when we toss the die.

b.) 9/216

c.) 9/216

d.) 212/216

Step-by-step explanation:

Sample space means all possible outcomes, since we know the dice are 6 face, then the sample space becomes all possible outcomes when we toss the die.

[1,1,1] [1,1,2] [1,1,3] [1,1,4] [1,1,5] [1,1,6]

[1,2,1] [1,2,2] [1,2,3] [1,2,4] [1,2,5] [1,2,6],

[1,3,1] [1,3,2] [1,3,3] [1,3,4] [1,3,5] [1,3,6]

[1,4,1] [1,4,2] [1,4,3] [1,4,4] [1,4,5] [1,4,6]

[1,5,1] [1,5,2] [1,5,3] [1,5,4] [1,5,5] [1,5,6]

[1,6,1] [1,6,2] [1,6,3] [1,6,4] [1,6,5] [1,6,6]

[2,1,1] [2,1,2] [2,1,3] [2,1,4] [2,1,5] [2,1,6]

[2,2,1] [2,2,2] [2,2,3] [2,2,4] [2,2,5] [2,2,6],

[2,3,1] [2,3,2] [2,3,3] [2,3,4] [2,3,5] [2,3,6]

[2,4,1] [2,4,2] [2,4,3] [2,4,4] [2,4,5] [2,4,6]

[2,5,1] [2,5,2] [2,5,3] [2,5,4] [2,5,5] [2,5,6]

[2,6,1] [2,6,2] [2,6,3] [2,6,4] [2,6,5] [2,6,6]

[3,1,1] [3,1,2] [3,1,3] [3,1,4] [3,1,5] [3,1,6]

[3,2,1] [3,2,2] [3,2,3] [3,2,4] [3,2,5] [3,2,6],

[3,3,1] [3,3,2] [3,3,3] [3,3,4] [3,3,5] [3,3,6]

[3,4,1] [3,4,2] [3,4,3] [3,4,4] [3,4,5] [3,4,6]

[3,5,1] [3,5,2] [3,5,3] [3,5,4] [3,5,5] [3,5,6]

[3,6,1] [3,6,2] [3,6,3] [3,6,4] [3,6,5] [3,6,6]

[4,1,1] [4,1,2] [4,1,3] [4,1,4] [4,1,5] [4,1,6]

[4,2,1] [4,2,2] [4,2,3] [4,2,4] [4,2,5] [4,2,6],

[4,3,1] [4,3,2] [4,3,3] [4,3,4] [4,3,5] [4,3,6]

[4,4,1] [4,4,2] [4,4,3] [4,4,4] [4,4,5] [4,4,6]

[4,5,1] [4,5,2] [4,5,3] [4,5,4] [4,5,5] [4,5,6]

[4,6,1] [4,6,2] [4,6,3] [4,6,4] [4,6,5] [4,6,6]

[5,1,1] [5,1,2] [5,1,3] [5,1,4] [5,1,5] [5,1,6]

[5,2,1] [5,2,2] [5,2,3] [5,2,4] [5,2,5] [5,2,6],

[5,3,1] [5,3,2] [5,3,3] [5,3,4] [5,3,5] [5,3,6]

[5,4,1] [5,4,2] [5,4,3] [5,4,4] [5,4,5] [5,4,6]

[5,5,1] [5,5,2] [5,5,3] [5,5,4] [5,5,5] [5,5,6]

[5,6,1] [5,6,2] [5,6,3] [5,6,4] [5,6,5] [5,6,6]

[6,1,1] [6,1,2] [6,1,3] [6,1,4] [6,1,5] [6,1,6]

[6,2,1] [6,2,2] [6,2,3] [6,2,4] [6,2,5] [6,2,6],

[6,3,1] [6,3,2] [6,3,3] [6,3,4] [6,3,5] [6,3,6]

[6,4,1] [6,4,2] [6,4,3] [6,4,4] [6,4,5] [6,4,6]

[6,5,1] [6,5,2] [6,5,3] [6,5,4] [6,5,5] [6,5,6]

[6,6,1] [6,6,2] [6,6,3] [6,6,4] [6,6,5] [6,6,6]

b.) Probability that the sum is 16 or more is

Pr[4,6,6] + pr[5,5,6] + pr [ 5,6,5] + pr [5,6,6] + pr [6,5,5] + pr [6,5,6] + pr [6,6,4] + pr[6,6,5] + pr [6,6,6]

Becomes:

[1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ = 9/216

Probability that the sum is 4 or 5 is

Pr [ 1,1,2] or pr[1,2,2] or pr [1,1,3] or pr [1,2,1] or pr[2,1,2] or pr[1,3,1] or pr[3,1,1] or pr[2,1,1] or pr[2,2,1]

Becomes:

[1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ = 9/216

Probability that the sum is less than 17

We take it as:

1- probability that the sum is 17 and above.

Now probability that the sum is 17 and above becomes

pr[5,6,6] or pr[6,5,6] or pr[6,6,5] or pr[6,6,6]

= [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ = 4/216

Hence, probability that the sum is less than 17 becomes:

1-4/216 = 212/216.

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3 years ago
A man tips a server $3.00 on meal cpsting $14.50. What percentage of this cost os the tip?
guajiro [1.7K]

Answer:

20.7%

Step-by-step explanation:

3/14.5 = 0.2068

Take 0.2068 and multiply it by 100 to get its percent value which is about 20.7%

$3.00 is 20.7% of $14.50

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3 years ago
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