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scZoUnD [109]
3 years ago
7

Explain how to find the difference between 9^10/9^11 and 9^11/9^10

Mathematics
2 answers:
kkurt [141]3 years ago
6 0

Answer:

jhv

Step-by-step explanation:

kjbb

levacccp [35]3 years ago
6 0

Answer:

-80/81

Step-by-step explanation:

(9^11)/(9^(10))=9^(11-10)=9

(9^10)/(9^(11))=9^(10-11)=9^(-1)=1/9

Difference is 1/9 - 9=-80/81

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Find the terms, coefficients, and constants of 4x+x+17
andreyandreev [35.5K]

Answer:

17 and x=constant,

4x=coefficient

hope this helps

have a good day :)

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
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
The head of a nail is circular, as shown. The head of this nail has a diameter of 6 millimeters.
Tcecarenko [31]

Answer:

its A

Step-by-step explanation:

Hope it helps

6 0
3 years ago
Victoria renovated a house for $71,000. If she paid 506,000 for the house, what is the percent of the original cost did she spen
castortr0y [4]

Answer: this is rounded out, and an estimate but...

Step-by-step explanation:

It appears to be 14.03% I could be wrong, but this was calculated so maybe not.

8 0
3 years ago
A box contains 48 tickets for six door prizes. Six tickets are drawn, and not replaced, to declare the winners. If serge purchas
Lyrx [107]
The tickets aren't replaced, so multiply the probabilities, decreasing the  denominator each time.

a) 1/48
b) (1/48)(1/47)
c) (1/48)(1/47)(1/46)(1/45)(1/44)(1/43)
6 0
3 years ago
Read 2 more answers
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