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Finger [1]
3 years ago
7

WILL GIVE BRAINLIEST!! ASAP Which of the following expressions are equivalent to 6^-3

Mathematics
1 answer:
olga_2 [115]3 years ago
7 0

Answer: 1/6^3 AND 1/216

Step-by-step explanation:

A negative number always goes to the denominator to become positive.

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1.2 ÷ 4 = 12 tenths ÷ 4
Harlamova29_29 [7]

Answer:

Yes you're right, it does

3 0
3 years ago
Please help pretty please.
Masteriza [31]

Answer:

reflection then translation

Step-by-step explanation:

you reflected of the y axis and then you translated 2 left and 5 up which brings you to 3

3 0
1 year ago
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
4 years ago
18.
chubhunter [2.5K]
A bakers dozen is 12 which would be $0.41 for each cookie. If you meant 13, it would be $0.38. All you have to do is divide 4.94 by the amount of cookies
7 0
3 years ago
The graph of h(x) is shown. Graph of h of x that begins in quadrant two and decreases rapidly following the vertical line which
fgiga [73]

Answer:

<u>x-intercept</u>

The point at which the curve <u>crosses the x-axis</u>, so when y = 0.

From inspection of the graph, the curve appears to cross the x-axis when x = -4, so the x-intercept is (-4, 0)

<u>y-intercept</u>

The point at which the curve <u>crosses the y-axis</u>, so when x = 0.

From inspection of the graph, the curve appears to cross the y-axis when y = -1, so the y-intercept is (0, -1)

<u>Asymptote</u>

A line which the curve gets <u>infinitely close</u> to, but <u>never touches</u>.

From inspection of the graph, the curve appears to get infinitely close to but never touches the vertical line at x = -5, so the vertical asymptote is x = -5

(Please note:  we cannot be sure that there is a horizontal asymptote at y = -2 without knowing the equation of the graph, or seeing a larger portion of the graph).

5 0
2 years ago
Read 2 more answers
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