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MArishka [77]
3 years ago
14

Find the area of the triangle.

Mathematics
2 answers:
Klio2033 [76]3 years ago
8 0

Answer:

<em>(D). 5 km²</em>

Step-by-step explanation:

QR² = 11² + 14² - 2(11)(14)cos77°

QR ≈ 15.739 km

P = 14 + 11 + 15.739 = 40.739 km

s = \frac{P}{2} ≈ 20.37 km

A_{PQR} = \sqrt{20.37(20.37-11)(20.37-14)(20.37-15.739)} ≈ <em>75 km²</em>

tiny-mole [99]3 years ago
4 0
The answer is B
The answer is B
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Mademuasel [1]

Answer:

13 is the answer ok I know but cheating in exam is bad

5 0
3 years ago
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Find m∠U.<br> Write your answer as an integer or as a decimal rounded to the nearest tenth.
olga2289 [7]

Answer:

Answer:

50.2

Step-by-step explanation:

ut =  \sqrt{ {6}^{2}  +  {5}^{2} }   \\  \sqrt{36 + 25}  \\  \sqrt{61 }  \\ using \: cosine \: rule \\ cos \: u =  \frac{ {5}^{2} + 61 - 36 }{2 \times 5 \times  \sqrt{61} }  \\  \\  =  \frac{25 + 61 - 36}{78}  \\  \\  = \frac{50}{78 }  \\  \cos \: u = 0.64 \\ m∠U. = 50.2

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2 years ago
Show that 1n^ 3 + 2n + 3n ^2 is divisible by 2 and 3 for all positive integers n.
Dmitry_Shevchenko [17]

Prove:

Using mathemetical induction:

P(n) = n^{3}+2n+3n^{2}

for n=1

P(n)  = 1^{3}+2(1)+3(1)^{2} = 6

It is divisible by 2 and 3

Now, for n=k, n > 0

P(k) = k^{3}+2k+3k^{2}

Assuming P(k) is divisible by 2 and 3:

Now, for n=k+1:

P(k+1) = (k+1)^{3}+2(k+1)+3(k+1)^{2}

P(k+1) = k^{3}+3k^{2}+3k+1+2k+2+3k^{2}+6k+3

P(k+1) = P(k)+3(k^{2}+3k+2)

Since, we assumed that P(k) is divisible by 2 and 3, therefore, P(k+1) is also

divisible by 2 and 3.

Hence, by mathematical induction, P(n) = n^{3}+2n+3n^{2} is divisible by 2 and 3 for all positive integer n.

3 0
3 years ago
43⋅(10025)−50 please help me someone
Mashutka [201]
43 * (10025) - 50 = 431075 - 50 = 431025
5 0
3 years ago
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Please help me !!!!!!!!!!!!
Law Incorporation [45]
∠A+∠B=∠ACD
So, ∠A=93-31=62°
I think it mean that, right?
8 0
3 years ago
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