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3 years ago

Find sin A and cos A

Mathematics

2 answers:

myrzilka [38]3 years ago

6 0

Answer:

Step-by-step explanation:

sinA=opp/hyp

sinA=16/20=4/5

cosA=adj/hyp

cosA=12/20=3/5

SVETLANKA909090 [29]3 years ago

4 0

Answer:

We found sinA= 3/5 and cosA=4/5

Option A is correct.

Step-by-step explanation:

The formula to find sin A is :

$sin A = \frac{Perpendicular}{Hypotenuse}$

while the formula to find cos A is:

$cosA=\frac{Base}{Hypotenuse}$

So, in the figure given:

Base = 16

Perpendicular = 12

Hypotenuse = 20

Now, finding sin A and cos A using formulas and above values

$sin A = \frac{Perpendicular}{Hypotenuse}\\sin A = \frac{12}{20}\\sinA=\frac{3}{5}$

$cosA=\frac{Base}{Hypotenuse}\\cosA=\frac{16}{20}\\cosA=\frac{4}{5}$

So, we found sinA= 3/5 and cosA=4/5

Option A is correct.

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Evaluate the function below for x = 4. F(x) = x3 + 2x2 + 1

pav-90 [236]

Answer:

Step-by-step explanation:

Plug in:

F(4) = (4)3 + 2(4)2 + 1

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3 years ago

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3 years ago

F(x) = lnx/x

My name is Ann [436]

$f(x) = \frac{lnx}{x}$

(a) $f'(x) = \frac{d}{dx}[\frac{lnx}{x}]$
Using the quotient rule:
$f'(x) = \frac{\frac{1}{x} \cdot x - lnx}{x^{2}}$
$f'(x) = \frac{1 - lnx}{x^{2}}$

For maximum, f'(x) = 0;
$\frac{1 - lnx}{x} = 0$
$lnx = 1, x = e$

(b) Deduce:
 $e^{x} \geq x^{e}, x > 0$

Soln: Since x = e is the greatest value, then f(e) ≥ f(x) > f(0)
$\frac{ln(e)}{e} \geq \frac{lnx}{x}$
$\frac{1}{e} \geq \frac{lnx}{x}$ , since ln(e) is simply equal to 1

Now, since x > 0, then we don't have to worry about flipping the signs when multiplying by x.
$\frac{x}{e} \geq lnx$
$x \geq elnx$
$x \geq ln(x^{e})$

Taking the exponential to both sides will cancel with the natural logarithmic function in the right hand side to produce:
$e^{x} \geq e^{ln(x^{e}})$
$e^{x} \geq x^{e}$ , as required.