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3 years ago

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If sec theta=1.4, find cos theta

1 answer:

Eduardwww [97]3 years ago

8 0

$\bf sec(\theta)=\cfrac{1}{cos(\theta)}\\\\
-----------------------------\\\\
sec(\theta)=1.4\implies \cfrac{1}{cos(\theta)}=1.4\implies \cfrac{1}{1.4}=cos(\theta)$

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a circle has a center at (-2,7). if a point on the circle has coordinates (3,-1) then which of the following would be the length

pickupchik [31]

Given that Circle O is centered at (3,-2) and the circle passesthroughR(7,1),Thedistance between O and R is the radius (r).r=√(7−3)2(1−(−2))2=√25=5diameter =2⋅r=2⋅5=10.

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Direct Variation: If y = -16, when x = 4, find x when y = -96.

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Answer:

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Step-by-step explanation:

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3 years ago

The daily average temperature in Santiago, Chile, varies over time in a periodic way that can be modeled approximately by a trig

natali 33 [55]

Answer:

a) the trigonometric function is;

$y = 7.5 sin ( \frac{2 \pi}{365}t + \frac{337 \pi}{730})+ 21.5$

b) $y = 28.36^0 \ C$ ( to two decimal places)

Step-by-step explanation:

This data can be represented by the sinusoidal function of the form :

$\mathbf{y = A sin (Bt -C)+D}$

where A = amplitude and which can be determined via the formula:

$A = \dfrac{largest \ temperature - lowest \ temperature}{2}$

$A = \dfrac{29-14}{2}$

$A = \dfrac{15}{2}$

A = 7.5° C

where B = the frequency;

Since the data covers a period of 3 days ; then $\dfrac{2 \pi}{B } =365$

$B = \dfrac{2 \pi}{365}$ ( where 365 is the time period )

The vertical shift is found by the equation D;

D = $\frac{largest \ temperature + lowest \ temperature}{2}$

D = $\frac{29+14}{2}$

D = 21.5

Replacing the values of A ; B and D into the above sinusoidal function; we have :

$y = 7.5 sin (\frac{2 \pi}{365}t -C) + 21.5$

From the question; when it is 7th of the year ( i.e January 7);

t = 7 and the temperature (y) = 29° C

replacing that too into the above equation; we have:

$29= 7.5 sin (\frac{2 \pi}{365}*7 -C) + 21.5$

$29= 7.5 sin (\frac{14 \pi}{365} -C) + 21.5$

$\frac{29-21.5}{7.5}= sin (\frac{14 \pi}{365} -C)$

$1= sin (\frac{14 \pi}{365} -C)$

$sin^{-1}(1)= (\frac{14 \pi}{365} -C)$

$\frac{\pi}{2}= (\frac{14 \pi}{365} -C)$

$C= (\frac{14 \pi}{365} -\frac{\pi}{2})$

$C= (\frac{28 \pi- 365 \pi}{730} )$

$C= \frac{-337 \pi}{730}$

Thus; the trigonometric function is;

$y = 7.5 sin ( \frac{2 \pi}{365}t + \frac{337 \pi}{730})+ 21.5$

Similarly; to determine the temperature o Jan 31; i.e when t= 31 ; we have :

$y = 7.5 sin ( \frac{2 \pi}{365}*31+ \frac{337 \pi}{730})+ 21.5$

$y = 7.5 sin ( \frac{62 \pi}{365}+ \frac{337 \pi}{730})+ 21.5$

$y = 7.5 sin ( \frac{124 \pi+ 337 \pi }{730})+ 21.5$

$y = 7.5 sin ( \frac{461 \pi }{730})+ 21.5$

$y = 7.5 *( 0.915)+ 21.5$

$y = 6.8689+ 21.5$

$y = 28.36^0 \ C$ ( to two decimal places)

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3 years ago

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