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solmaris [256]
3 years ago
7

mrs.baker paid $2.50 for 5 pounds of bannas.write an equation relating the cost c to the number of pounds p of bannas.how much w

ould mrs.baker pay for 8 pounds of bannanas?
Mathematics
1 answer:
Brums [2.3K]3 years ago
8 0
So, if she paid 2.50 for 5 pounds, how much would she pay for one pound? we divide by 5:

2.50/5=0.50

so one pound costs 0.50 dollars.

then for 8 pounds, she'd pay eight times this much, that is 4 dollars.
 
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The sum of three consecutive numbers is 72 what is the smallest of the numbers
mylen [45]

Answer:

23

Step-by-step explanation:

Suppose,

the first number = x

the second number = x+1

the third number = x+1+1 = x+2

Question wise,

x+x+1+x+2 = 72

3x+3 = 72

3x = 72-3 = 69

x = 69/3 = 23

so the numbers are,

23,(23+1)=24 and (23+2)=25

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I need help with #11
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\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
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Answer:

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