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elena55 [62]
3 years ago
11

find the missing side. round to the nearest tenth. use pythagorean theorem to find the third side length​

Mathematics
1 answer:
emmasim [6.3K]3 years ago
4 0

Answer:

x is tangent of 39 = x/ 19, tangent of 39 is 3. 615...... then 19 multiply by 3.615....., x would get 68. 68

Step-by-step explanation:

tangent is opposite over adjacent then put all your numbers in to the formula.

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Solve The Triangle (See Picture!!)
joja [24]

Answer:

Step-by-step explanation:

use sin formula

\frac{a}{sin \alpha } =\frac{b}{sin \beta } =\frac{c}{sin \gamma} \\\frac{10}{sin~40} =\frac{12}{sin ~\beta } \\sin~\beta =\frac{12}{10} \times sin~40\\ \beta=sin^{-1}(1.2 sin ~40)\approx 50.5 ^\circ\\\gamma=180-(40+50.5)=89.5^\circ\\\frac{c}{sin~89.5}=\frac{10}{sin~40}  \\c=\frac{sin~89.5}{sin ~40} \times 10 \approx 15.6\\D

4 0
3 years ago
#18 There are 8 students
S_A_V [24]

There are 20 students in the class.

8 0
3 years ago
What is he diameter of a circle with a circumference of 235.5 cm
GalinKa [24]
Diameter of a circle = \frac{Circumference}{ \pi }
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Diameter of a circle 
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6 0
3 years ago
NO LINKS!!! Please help fill in the blanks. Part 11a
Lorico [155]

Answer:

  see attached

Step-by-step explanation:

Here's your worksheet with the blanks filled.

__

Of course, you know these log relations:

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4 0
2 years ago
Please calculate this limit <br>please help me​
Tasya [4]

Answer:

We want to find:

\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n}

Here we can use Stirling's approximation, which says that for large values of n, we get:

n! = \sqrt{2*\pi*n} *(\frac{n}{e} )^n

Because here we are taking the limit when n tends to infinity, we can use this approximation.

Then we get.

\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n} = \lim_{n \to \infty} \frac{\sqrt[n]{\sqrt{2*\pi*n} *(\frac{n}{e} )^n} }{n} =  \lim_{n \to \infty} \frac{n}{e*n} *\sqrt[2*n]{2*\pi*n}

Now we can just simplify this, so we get:

\lim_{n \to \infty} \frac{1}{e} *\sqrt[2*n]{2*\pi*n} \\

And we can rewrite it as:

\lim_{n \to \infty} \frac{1}{e} *(2*\pi*n)^{1/2n}

The important part here is the exponent, as n tends to infinite, the exponent tends to zero.

Thus:

\lim_{n \to \infty} \frac{1}{e} *(2*\pi*n)^{1/2n} = \frac{1}{e}*1 = \frac{1}{e}

7 0
3 years ago
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