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ryzh [129]
3 years ago
5

What is the angle (t) if

sqrt{65}/65" alt="sin(t)=4\sqrt{65}/65" align="absmiddle" class="latex-formula">. Recall that theta lies in quadrant IV. The answer to this is 330.3 degrees, but please show how that answer is found.

Mathematics
1 answer:
dybincka [34]3 years ago
3 0

Answer:

330.3 degrees

Step-by-step explanation:

sin t = 4√65 / 65 reduces to sin t = 4 / √65.  As the sine function is defined as   (opp side) / (hypotenuse), we see that opp side = 4 and hyp = √65 must be true.  But sin t is positive in Quadrants I and II, not in Quadrant IV.

I will take the liberty of assuming you meant  sin t = -4 / √65.

Then (opp side) = -4 and (hyp) = √65.

Use the inverse sine function on a calculator to determine this angle t:

-arcsin(4/√65) comes to -arcsin 0.4961 = -0.5191 radians.

Converting -0.5191 radians to degree measure results in

t = 360° - 29.745°, or 330.2551 degrees.  This rounds off to 330.3 degrees.

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Given that f(x) = 6x - 5 g(x) = 3x + 4 and h(x) = 4x – 6
bazaltina [42]

Answer:

(i) g(-2)=-2

(ii) g[h(x)]=12x-14

(iii) f[g(2)]=55

(iv) (g\circ h)(2)=10

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Step-by-step explanation:

The given functions are

f(x)=6x-5

g(x)=3x+4

h(x)=4x-6

(i) Find g(-2).

Substitute x=-2 in g(x).

g(-2)=3(-2)+4\Rightarrow -6+4=-2

(ii) Find g[h(x)]

g[h(x)]=g(4x-6)                 (h(x)=4x-6)

g[h(x)]=3(4x-6)+4           (g(x)=3x+4)

g[h(x)]=12x-18+4

g[h(x)]=12x-14

(iii) Find f[g(2)].

f[g(2)]=f[3(2)+4]           (g(x)=3x+4)

f[g(2)]=f(6+4)

f[g(2)]=f(10)

f[g(2)]=6(10)-5           f(x)=6x-5

f[g(2)]=55

iv) Find gᴏh(2).

(g\circ h)(2)=g[h(2)]

(g\circ h)(2)=12(2)-14           (From part (ii) we get g[h(x)]=12x-14)

(g\circ h)(2)=10

(v) Find h^{-1}(11)

First find h^{-1}(x).

h(x)=4x-6

y=4x-6

x=4y-6

x+6=4y

Divide both sides by 4.

\frac{x+6}{4}=y

h^{-1}(x)=\frac{x+6}{4}

Substitute x=11 in the inverse function.

h^{-1}(11)=\frac{11+6}{4}

h^{-1}(11)=\frac{17}{4}

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