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ZanzabumX [31]
4 years ago
10

Determine the quadrant when the terminal side of the angle lies according to the following conditions: sin (t) > 0, cos (t) &

lt; 0.
Mathematics
1 answer:
grin007 [14]4 years ago
7 0
Check the problem again!Cosecant (csc) is the reciprocal of sine (sin) so they are always either both positive or both negative. Perhaps it should be sine < 0 and cosine > 0? In that case, sine is less than zero in quadrants 3 and 4, and cosine is greater than zero in quadrants 1 and 4, so this angle can only lie in quadrant 4. On the unit circle, remember that cosine is the x-coordinate of the terminal side of the angle and sine is the y-coordinate.  Quadrant 1 is that where both sine and cosine are greater than zero.  The rest of them are numbered consecutively going counter-clockwise; so quadrant 2 has cos < 0 and sin > 0, quadrant 3 has cos < 0 and sin < 0, and quadrant 4 has cos > 0 and sin < 0
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Please help me out !!
Dmitrij [34]
<h3>Solution:</h3>

\huge \pink {\tt {{128}^{ \frac{3}{7} }  =  \sqrt[7]{ {128}^{3} }  =  {2}^{3}  }}

\huge \pink {\tt {= 8}}

<h2>Answer:</h2>

\huge \purple {\tt {8 \: is \: the \: answer}}

8 0
3 years ago
I don’t really understand this can some help me with the right answer
MatroZZZ [7]
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7 0
3 years ago
Mariela bought gas for her car. Gas costs $2.12 per gallon, and
Montano1993 [528]

Answer:

$49.82

Step-by-step explanation:

Since each gallon is $2.12 you would just do $2.12 times 23.5 which equals $49.82. So Mariela spent $49.82 on gas

5 0
3 years ago
Read 2 more answers
If hexagon JGHFKI Is congruent to hexagon SRTQUV, which pair of angles must be congruent?
pav-90 [236]

Angles J and S or I and V would be congruent because they are corresponding.

3 0
4 years ago
Read 2 more answers
Given f(x) =17 -x2 what is the average rate of change in f(x) over the interval [1,5]
Artist 52 [7]
By definition <span>the average rate of change is given by:
 </span>AVR =  \frac{f(x2) - f(x1)}{x2-x1}
 <span>We have the following function:
 </span>f(x) =17 - x^2
 We evaluate the function for the given interval:
 For X = 1:
 f(1) =17 - 1^2
 f(1) =17 - 1
 f(1) =16
 For X = 5:
 f(5) =17 - 5^2
 f(5) =17 - 25
 f(5) =-8
 Then, replacing values we have:
 AVR = \frac{-8 - 16}{5-1}
 AVR = \frac{-24}{4}
 AVR = -6
 Answer:
 <span>the average rate of change in f(x) over the interval [1,5] is:
 </span>AVR = -6


6 0
3 years ago
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