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Vinil7 [7]
3 years ago
7

a stop sign is shown. each side measures 12.4 inches , and the distance from any side to the opposite side is 30 inches

Mathematics
1 answer:
Serhud [2]3 years ago
7 0
2332 inches from the opposite side
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Choose an angle or point from the preimage and name its image
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From the preimage, the angle or point I choose is P and I name its image as: P maps to P’.

 

The set of all elements of the domain that map to the members of S is the inverse image or preimage of a particular subset S of the codomain of a function.

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What is the distance from M(9,-4) to NC-1, -2)?
Ilia_Sergeevich [38]

Answer:

Exact Form: 2√26

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Step-by-step explanation:

8 0
3 years ago
What is the area of the figure?
Pie

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80.2

Step-by-step explanation:

6 0
2 years ago
8. Find all the real fourth roots of 256
mezya [45]
X^4=256
x^4-256=0
(x^2)^2-(16)^2=0
(x^2+16)(x^2-16)=0
either x^2+16=0
it gives complex roots.
or x^2-16=0
or x^2-4^2=0
(x+4)(x-4)=0
either x+4=0,x=-4
or x-4=0
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7 0
3 years ago
Write the equation of a hyperbola with vertices (0, -4) and (0, 4) and foci (0, -5) and (0, 5).
andre [41]
Check the picture below.  So, more or less looks like so.

notice, the center is clearly at the origin, and notice how long the "a" component is, also, bear in mind that, is opening towards the y-axis, that means the fraction with the "y" variable is the positive one.

Also notice, the "c" distance from the center to either foci, is just 5 units.

\bf \textit{hyperbolas, vertical traverse axis }\\\\
\cfrac{(y-{{ k}})^2}{{{ a}}^2}-\cfrac{(x-{{ h}})^2}{{{ b}}^2}=1
\qquad 
\begin{cases}
center\ ({{ h}},{{ k}})\\
vertices\ ({{ h}}, {{ k}}\pm a)\\
c=\textit{distance from}\\
\qquad \textit{center to foci}\\
\qquad \sqrt{{{ a }}^2+{{ b }}^2}
\end{cases}\\\\
-------------------------------\\\\

\bf \begin{cases}
h=0\\
k=0\\
a=4\\
c=5
\end{cases}\implies \cfrac{(y-{{ 0}})^2}{{{ 4}}^2}-\cfrac{(x-{{ 0}})^2}{{{ b}}^2}=1\implies \cfrac{y^2}{16}-\cfrac{x^2}{b^2}=1
\\\\\\
c=\sqrt{a^2+b^2}\implies c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b
\\\\\\
\sqrt{5^2-4^2}=b\implies \boxed{3=b}
\\\\\\
\cfrac{y^2}{16}-\cfrac{x^2}{3^2}=1\implies \boxed{\cfrac{y^2}{16}-\cfrac{x^2}{9}=1}

7 0
3 years ago
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