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WITCHER [35]
3 years ago
12

How are the locations on the coordinate grid different for the ordered pairs (7,0) and (0,7)

Mathematics
2 answers:
alex41 [277]3 years ago
8 0
They will be on different axis.

The 7 in (7,0) will first go right 7 times and stay on that line. In (0,7) there will not be a point on the x axis.

ladessa [460]3 years ago
7 0
They would both be on a different axis.
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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}
3 0
3 years ago
What is the product? (-2a^2+a)(5a^2-6s)
ANEK [815]
The answer would be A
6 0
3 years ago
Sumas y restas w+y=9 3w-y=11
Westkost [7]

Answer:

w = 5

y = 4

Step-by-step explanation:

w+y=9

3w-y=11

4w = 20

w = 5

y = 4

3 0
3 years ago
How do I solve this triangle?
svetlana [45]

Answer:

check the length and the angles and check the height of it

7 0
3 years ago
A pair of shoes cost $88. You get
Maurinko [17]

Answer:

$77.44

Step-by-step explanation:

If the shoes cost $88 but you get a 20% discount, subtract 20% from 88: -->

$88-20% = $70.4

Then add 8% of 88 to 70.4 -->

8% of $88 = $7.04.

$70.4 + $7.04 = $77.44

Therefore the final price of the shoes are $77.44

Hope this helps!!!~~ <3

7 0
3 years ago
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