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zaharov [31]
3 years ago
13

I need help with 3 B please!

Mathematics
1 answer:
Lostsunrise [7]3 years ago
5 0

Answer

im sorry i tried to help it wouldnt load

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Nat2105 [25]
The unit rate is 2/3 because in the line y=2/3x, 2/3 is the slope (y=mx+b). The slope of the line is pretty much the definition of a unit rate because the unit rate is a constant addition or subtraction over and over. In addition, it is a straight line(y=mx+b) and the values are going up at a constant rate of 2/3.
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7 2 − 140 degrees for angles
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What?

Step-by-step explanation:

6 0
2 years ago
10 points, please help me and explain how to do this with answers!
8_murik_8 [283]
\bf f(x)=log\left( \cfrac{x}{8} \right)\\\\
-----------------------------\\\\
\textit{x-intercept, setting f(x)=0}
\\\\
0=log\left( \cfrac{x}{8} \right)\implies 0=log(x)-log(8)\implies log(8)=log(x)
\\\\
8=x\\\\
-----------------------------

\bf \textit{y-intercept, is setting x=0}\\
\textit{wait just a second!, a logarithm never gives 0}
\\\\
log_{{  a}}{{  b}}=y \iff {{  a}}^y={{  b}}\qquad\qquad 
%  exponential notation 2nd form
{{  a}}^y={{  b}}\iff log_{{  a}}{{  b}}=y 
\\\\
\textit{now, what exponent for "a" can give  you a zero? none}\\
\textit{so, there's no y-intercept, because "x" is never 0 in }\frac{x}{8}\\
\textit{that will make the fraction to 0, and a}\\
\textit{logarithm will never give that, 0 or a negative}\\\\


\bf -----------------------------\\\\
domain
\\\\
\textit{since whatever value "x" is, cannot make the fraction}\\
\textit{negative or become 0, , then the domain is }x\ \textgreater \ 0\\\\
-----------------------------\\\\
range
\\\\
\textit{those values for "x", will spit out, pretty much}\\
\textit{any "y", including negative exponents, thus}\\
\textit{range is }(-\infty,+\infty)
 p, li { white-space: pre-wrap; }

----------------------------------------------------------------------------------------------




now on 2)

\bf f(x)=\cfrac{3}{x^4}   if the denominator has a higher degree than the numerator, the horizontal asymptote is y = 0, or the x-axis,

in this case, the numerator has a degree of 0, the denominator has 4, thus y = 0


vertical asymptotes occur when the denominator is 0, that is, when the fraction becomes undefined, and for this one, that occurs at  x^4=0\implies x=0  or the y-axis

----------------------------------------------------------------------------------------------


now on 3)

\bf f(x)=\cfrac{1}{x}


now, let's see some transformations templates

\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}}
\end{array}


\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks horizontally by  } {{  A}}\cdot {{  B}}\\
\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}\\
\bullet \textit{ vertical shift by }{{  D}}\\
\qquad if\ {{  D}}\textit{ is negative, downwards}\\
\qquad if\ {{  D}}\textit{ is positive, upwards}
\end{array}


now, let's take a peek at g(x)

\bf \begin{array}{lcllll}
g(x)=&-&\cfrac{1}{x}&+3\\
&\uparrow &&\uparrow \\
&\textit{upside down}&&
\begin{array}{llll}
\textit{vertical shift up}\\
\textit{by 3 units}
\end{array}
\end{array}


3 0
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15r^2 - 14r -12 is the correct answer
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(56 + 8 x 2) divided by 9
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Answer:8

Step-by-step explanation:

(56+8 x 2) divided by 9

(56+16) divided by 9

72/9

8

8 0
3 years ago
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