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dezoksy [38]
2 years ago
8

Answer this question pls? Its 20 pts

Mathematics
2 answers:
ICE Princess25 [194]2 years ago
6 0

Answer:

B   a = 180 -77-60

Step-by-step explanation:

The angles add to 180 degrees since they form a straight line

77+ 60+a = 180

a = 180 -77-60

MArishka [77]2 years ago
4 0

The answer is B.

Step-by-step explanation:

We know that the sum of angles in a straight line = 180°

Therefore:

77° + 60° + a = 180°

a = 180° - 77° - 60°

Hence, option B is the answer.

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ioda

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C an urban planner

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Anita has a cellphone contract that costs her R100 per month plus 85 cents per peak time SMS,and 25 cents per off-peak time SMS.
algol [13]

Answer:

Anita's monthly bill will be $164.25.

Step-by-step explanation:

Since Anita has a cellphone contract that costs her R100 per month plus 85 cents per peak time SMS, and 25 cents per off-peak time SMS, if she sends 45 SMSs during peak time and 105 SMSs during off-peak time In a month To determine what will her monthly bill come to, the following calculation must be performed:

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Therefore, her monthly bill will be $ 164.25.

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3 years ago
Simplify ( root 3 + root 7) over 2​
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5 r o 2 t

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(√3+√7)²

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Plz help me fast Simplify t12/t6
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t^6

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Find the remaining trigonometric ratios of θ if csc(θ) = -6 and cos(θ) is positive
VikaD [51]
Now, the cosecant of θ is -6, or namely -6/1.

however, the cosecant is really the hypotenuse/opposite, but the hypotenuse is never negative, since is just a distance unit from the center of the circle, so in the fraction -6/1, the negative must be the 1, or 6/-1 then.

we know the cosine is positive, and we know the opposite side is -1, or negative, the only happens in the IV quadrant, so θ is in the IV quadrant, now

\bf csc(\theta)=-6\implies csc(\theta)=\cfrac{\stackrel{hypotenuse}{6}}{\stackrel{opposite}{-1}}\impliedby \textit{let's find the \underline{adjacent side}}
\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
\pm\sqrt{6^2-(-1)^2}=a\implies \pm\sqrt{35}=a\implies \stackrel{IV~quadrant}{+\sqrt{35}=a}

recall that 

\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad\qquad 
cos(\theta)=\cfrac{adjacent}{hypotenuse}
\\\\\\
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}
\qquad \qquad 
% cotangent
cot(\theta)=\cfrac{adjacent}{opposite}
\\\\\\
% cosecant
csc(\theta)=\cfrac{hypotenuse}{opposite}
\qquad \qquad 
% secant
sec(\theta)=\cfrac{hypotenuse}{adjacent}

therefore, let's just plug that on the remaining ones,

\bf sin(\theta)=\cfrac{-1}{6}
\qquad\qquad 
cos(\theta)=\cfrac{\sqrt{35}}{6}
\\\\\\
% tangent
tan(\theta)=\cfrac{-1}{\sqrt{35}}
\qquad \qquad 
% cotangent
cot(\theta)=\cfrac{\sqrt{35}}{1}
\\\\\\
sec(\theta)=\cfrac{6}{\sqrt{35}}

now, let's rationalize the denominator on tangent and secant,

\bf tan(\theta)=\cfrac{-1}{\sqrt{35}}\implies \cfrac{-1}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{-\sqrt{35}}{(\sqrt{35})^2}\implies -\cfrac{\sqrt{35}}{35}
\\\\\\
sec(\theta)=\cfrac{6}{\sqrt{35}}\implies \cfrac{6}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{6\sqrt{35}}{(\sqrt{35})^2}\implies \cfrac{6\sqrt{35}}{35}
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3 years ago
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