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Vaselesa [24]
3 years ago
6

Your model train has one engine and eight train cars. Find the total number of ways you can arrange the train.

Mathematics
1 answer:
baherus [9]3 years ago
4 0
The answer is five! Don't tell me how I got that because my sister did the problem 2 minutes ago because I told her this question! Hope this help!
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How do I solve for y in the equation 2(3/4)+y=4?
yuradex [85]
2* .75 + y = 4
1.5 + y = 4
Subtract 1.5 From Each Side
Y = 2.5 Or 2 1/2
5 0
3 years ago
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When constructing an inscribed circle, which point is the center of the circle?
lesantik [10]

Answer:

  D.  The intersection of the angle bisectors.

Step-by-step explanation:

The incenter is equidistant from each of the sides of the triangle. An angle bisector is the set of points equidistant from the sides, so the incenter is the place where the angle bisectors meet.

6 0
3 years 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
3 years ago
Solve the system of equations.
kotegsom [21]
Y =  2.
x =  0?
i believe I am correct about x but I am not sure.
4 0
3 years ago
Read 2 more answers
What conic section degenerates into a line?
andrew11 [14]
<h3>Answer: A) parabola</h3>

Some degenerate parabola cases form a single straight line, while other cases form one pair of parallel lines.

A degenerate hyperbola forms two lines that intersect at the vertex of the cone. We can rule out choice B.

A degenerate circle is a single point, so we can rule out choice C.

A degenerate ellipse is also a single point. Any circle is an ellipse (but not the other way around). We can rule out choice D.

5 0
3 years ago
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