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Len [333]
3 years ago
9

Find all solutions of the equation in the interval [0,2π) sec(theta) +2=0

Mathematics
1 answer:
Komok [63]3 years ago
6 0
I will first reveal the most obvious secret of trig: Most of the problems are 30/60/90 or 45/45/90 triangles of some sort.

\sec \theta + 2  = 0

\dfrac{1 }{ \cos \theta} = -2

\cos \theta = - \frac 1 2

The rule to remember is \cos x = \cos a has solutions

x = \pm a + 2 \pi k \quad integer k

Continuing where we left off, a cosine of -1/2 is a 30/60/90 triangle in the second or third quadrant; we pick second, and a little thought tells us the angle is  120^\circ.

\cos \theta = - \frac 1 2

\cos \theta = \cos \frac {2\pi} 3

\theta = \pm \frac {2\pi} 3 + 2 \pi k \quad integer k

In the range we want, that's k=0 with the plus sign and k=1 with the minus, so 

\theta = \frac {2\pi} 3 or -\frac {2\pi} 3 + 2 \pi = \frac{4 \pi} 3

\theta = \frac {2\pi} 3 or \frac{4 \pi} 3
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I need help with number 2 please.
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At a bargain store.Tanya bought 5 items that each cost the same amount. Tony bought 6 items that each cost the same amount, but
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(A) For x representing the cost of one of Tanya's items, her total purchase cost 5x. The cost of one of Tony's items is then (x-1.75) and the total of Tony's purchase is 6(x-1.75). The problem statement tells us these are equal values. Your equation is ...

... 5x = 6(x -1.75)

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... 6(x -1.75) = 6·8.75 = 52.50 . . . . . the two purchases are the same value

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Please assist me with this problem​
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Answer:

  d.  about 50 times larger

Step-by-step explanation:

The given expressions for magnitude (M) can be solved for the intensity (I). Then the ratio of intensities is ...

  \dfrac{I_2}{I_1}=\dfrac{I_0\cdot 10^{M_2}}{I_0\cdot 10^{M_1}}=10^{M_2-M_1}=10^{4.2-2.5}\\\\=10^{1.7}\approx 50

The larger earthquake had about 50 times the intensity of the smaller one.

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3 years ago
SoMeonE plZ hAlp me lol
jeka94

C.√93


A=17

B=14

...........................

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3 years ago
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