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Vesnalui [34]
3 years ago
11

I really need help I have 20 minutes

Mathematics
2 answers:
nlexa [21]3 years ago
7 0

Answer:

a. -1.7

b. -1.2

c. -0.5

d. 0.6

e. 1.1

Step-by-step explanation:

Ede4ka [16]3 years ago
6 0

Answer:

bjgkhagnakjkmk knakhjkjugrfhewj uwgewguwghnrujwbewfb h bjbkejf

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If sin Θ = negative square root 3 over 2 and π < Θ < 3 pi over 2, what are the values of cos Θ and tan Θ? cos Θ = negative
andreev551 [17]

Answer:

The correct option is the option with:

cos Θ = 1/2

tan Θ = -√3

Step-by-step explanation:

Given that

sin Θ = -√3/2

We want to find the values of

cos Θ and tan Θ

First of all,

arcsin (-60) = -√3/2

=> Θ = 60

tan Θ = (sin Θ)/(cos Θ)

tan Θ = (-√3/2)/(cos Θ)

cos Θ tan Θ = (1/2)(-√3)

Knowing that Θ = -60,

and cos Θ = cos(-Θ), comparing the last equation, we have

cos Θ = 1/2

tan Θ = -√3

8 0
3 years ago
Read 2 more answers
What is the quotient for 2 8/9 / (-5 8/9)
AveGali [126]
The answer is 2764 if I’m not mistaken
3 0
3 years ago
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Which decimal number is less than 1.56?
Digiron [165]

The decimal number is 0.9

3 0
3 years ago
Help asap please,i don’t get the question
mario62 [17]

Answer:

see below

Step-by-step explanation:

I do not know what the 'given expression' is, bu the three UN-highlighted expressions in each question 1 and 2  are equivalent

  Only the highlighted one is not equiv to the other three

3 0
2 years ago
A circle has the center of (1,-5) and a radius of 5 determine the location of the point (4,-1)
Sliva [168]

"determine the location" or namely, is it inside the circle, outside the circle, or right ON the circle?

well, we know the center is at (1,-5) and it has a radius of 5, so the distance from the center to any point on the circle will just be 5, now if (4,-1) is less than that away, is inside, if more than that is outiside and if it's exactly 5 is right ON the circle.

well, we can check by simply getting the distance from the center to the point (4,-1).

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ \stackrel{center}{(\stackrel{x_1}{1}~,~\stackrel{y_1}{-5})}\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{-1})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d = \sqrt{[4-1]^2+[-1-(-5)]^2}\implies d=\sqrt{(4-1)^2+(-1+5)^2} \\\\\\ d = \sqrt{3^2+4^2}\implies d =\sqrt{9+16}\implies d=\sqrt{25}\implies \stackrel{\textit{right on the circle}}{d = 5}

5 0
3 years ago
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