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grigory [225]
3 years ago
6

Determine whether the given segments have the same length. Drag and drop each number

Mathematics
1 answer:
s344n2d4d5 [400]3 years ago
6 0

Answer:

Step-by-step explanation:

Length of a segment between two points (x_1,y_1) and (x_2,y_2) is given by the formula,

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Coordinates of extreme ends of BC and EF are,

B(1, 4), C(3, -2), E(-3, -2) and F(-1, 1)

Length of BC = \sqrt{(1-3)^2+(4+2)^2}

                      = \sqrt{4+36}

                      = 2\sqrt{10} ≈ 6.32 units

Length of EF = \sqrt{(-3+1)^2+(-2-1)^2}

                     = \sqrt{4+9}

                     = \sqrt{13} ≈ 3.61 units

The length of BC is 6.32. The length of EF is 3.61.So, BC and EF don't have the same length.

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  h^{-1}(x)\qquad\text{is found from }x=h(y)\\\\x=3y+8\\x-8=3y\\\\y=\dfrac{x-8}{3}\\\\\boxed{h^{-1}(x)=\dfrac{x-8}{3}}

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Use the graph of f(x) below to estimate the value of f '(3):
klio [65]
Well, looking at the graph, notice, its vertex is at 0,9, and it passes through the points -3,0 and 3,0, so hmm let's use 3,0, to get the equation of f(x)

\bf ~~~~~~\textit{parabola vertex form}
\\\\
\begin{array}{llll}
\boxed{y=a(x- h)^2+ k}\\\\
x=a(y- k)^2+ h
\end{array}
\qquad\qquad
vertex~~(\stackrel{}{ h},\stackrel{}{ k})\\\\
-------------------------------

\bf vertex~(0,9)~
\begin{cases}
h=0\\
k=9
\end{cases}\implies y=a(x-0)^2+9
\\\\\\
\textit{we also know that }
\begin{cases}
x=3\\
y=0
\end{cases}\implies 0=a(3-0)^2+9
\\\\\\
-9=9a\implies \cfrac{-9}{9}=a\implies -1=a\quad thus\quad \boxed{y=-x^2+9}
\\\\\\
\left. \cfrac{dy}{dx}=-2x \right|_{x=3}\implies -6
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4 years ago
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