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3 years ago

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Si sen A = 7/25 donde "A" pertenece al segundo cuadrante y el cos B = -4/5 donde "B" pertenece al segundo cuadrante Calcular cos

ec(A).cos(B).tan(A).cot(B)

1 answer:

lesya [120]3 years ago

3 0

Not sure just need answers because I don’t no the answers

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gretchen and henrik are meeting at a park. gretchen lives 2.4 miles away from the park and bikes at a constant rate of 12 miles

shepuryov [24]

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2)t=S/U=8.4/36=0,23
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What is the value of x?

Phantasy [73]

Answer:

value of x = 5.8 mm

Step-by-step explanation:

We have given,

Two right triangles EDH and EDG.

In right triangle EDH, EH = 56mm , DH = 35 mm

Using Pythagoras theorem we can find ED.

i.e EH² = ED²+DH²

56²=ED²+35²

ED²=56²-35²

ED = √(56²-35²) = 7√39 = 43.71 mm

Now, Consider right triangle EDG

Here, EG=44.8mm , GD = x+4 and ED = 7√39

Again using Pythagoras theorem,

EG² = ED² + DG²

44.8²= (7√39)²+ (x+4)²

(x+4)² = 44.8² - (7√39)²

x+4 = √(44.8² - (7√39)²)

x+4 = 9.8

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4 years ago

Sin(60-theta)sin(60+theta)

ehidna [41]

Step-by-step explanation:

$\sin(60 - \theta) \sin(60 + \theta) \\ = \{ \sin(60) \cos( \theta) - \sin( \theta) \cos(60) \} \{ \sin(60) \cos( \theta) + \cos(60) \sin( \theta) \} \\ = \{ \frac{ \sqrt{3} }{2} \cos( \theta) - \frac{1}{2} \sin( \theta) \} \{ \frac{ \sqrt{3} }{2} \cos( \theta) + \frac{1}{2} \sin( \theta) \} \\ \\$

from difference of two squares:

${ \boxed{(a - b)(a + b) = ( {a}^{2} - {b}^{2} ) }}$

therefore:

$= \{ {( \frac{ \sqrt{3} }{2}) }^{2} { \cos }^{2} \theta \} - \{ {( \frac{ \sqrt{3} }{2} )}^{2} { \sin }^{2} \theta \} \\ \\ = \frac{3}{4} { \cos }^{2} \theta - \frac{3}{4} { \sin}^{2} \theta$

factorise out ¾ :

$= \frac{3}{4} ( { \cos }^{2} \theta - { \sin}^{2} \theta) \\ \\ = { \boxed{ \frac{3}{4} \cos(2 \theta) }}$

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Alisiya [41]

The decimal is 0.25.

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4 years ago

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ludmilkaskok [199]

Answer:

Blue splat

Step-by-step explanation:

Blue splat

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