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Shkiper50 [21]
3 years ago
13

Please help I’m down bad atm

Mathematics
1 answer:
kogti [31]3 years ago
5 0

Answer:

the answer is an-as-1+1;a,=1

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In the diagram below, is the perpendicular bisector of JK. Which of the following statements must be true?
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GrogVix [38]

Answer:

i have no clue

Step-by-step explanation:


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I need help with #11
Troyanec [42]
\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\

\begin{array}{rllll} 
% left side templates
f(x)=&{{  A}}({{  B}}x+{{  C}})+{{  D}}
\\ \quad \\
y=&{{  A}}({{  B}}x+{{  C}})+{{  D}}
\\ \quad \\
f(x)=&{{  A}}\sqrt{{{  B}}x+{{  C}}}+{{  D}}
\\ \quad \\
f(x)=&{{  A}}(\mathbb{R})^{{{  B}}x+{{  C}}}+{{  D}}
\\ \quad \\
f(x)=&{{  A}} sin\left({{ B }}x+{{  C}}  \right)+{{  D}}
\end{array}

\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks horizontally by  } {{  A}}\cdot {{  B}}\\\\
\bullet \textit{ flips it upside-down if }{{  A}}\textit{ is negative}
\\\\
\bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\
\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\\\
\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\\\
\end{array}

\bf \begin{array}{llll}


\bullet \textit{ vertical shift by }{{  D}}\\
\qquad if\ {{  D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{  D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{  B}}}
\end{array}

now, with that template in mind, let's take a peek at this function

\bf \begin{array}{lllcclll}
y=&2(&1x&-2)^2&-4\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}\\\\
-----------------------------\\\\
A\cdot B=2\impliedby \textit{shrunk by a factor of 2, of half-size}\\\\
\cfrac{C}{B}= \cfrac{-2}{1}\implies -2\impliedby \textit{horizontal right shift of 2 units}\\\\
D=-4\impliedby \textit{vertical down shift of 4 units}

so, the graph of y=2(x-2)²-4, is really the same graph of y=x², BUT, narrower, and moved about horizontally and vertically
8 0
3 years ago
Read 2 more answers
What is the equation that represents the line?
Kryger [21]

Answer:

see explanation

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Calculate m using the slope formula

m = (y₂ - y₁ ) / (x₂ - x₁ )

with (x₁, y₁ ) = (- 2, - 6) and (x₂, y₂ ) = (2, - 3) ← 2 points on the line

m = \frac{-3+6}{2+2} = \frac{3}{4}, thus

y = \frac{3}{4} x + c ← is the partial equation

To find c substitute either ot the 2 points into the partial equation

Using (2, - 3), then

- 3 = \frac{3}{2} + c ⇒ c = - 3 - \frac{3}{2} = - \frac{9}{2}

y = \frac{3}{4} x - \frac{9}{2} ← equation of line

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3 years ago
Please help I don't understand i'm in the lowest class?
Anna71 [15]

Answer:

Step-by-step explanation:

6 0
2 years ago
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