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AVprozaik [17]
3 years ago
15

Whats 2+2 as a fraction

Mathematics
2 answers:
notka56 [123]3 years ago
8 0
2 + 2 = 4

4 / 1 = 4 

so the fraction form of " 2+2 (or 4) " would be " 4/1 "
Serga [27]3 years ago
6 0
[(2+2)]
--------- = 4/1
1
is how you solve it
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Jakes has $20 to spend on notebooks and pencils the notebooks and pencils. the notebooks cost 3.25 and the pencils cost $.50 wha
Leokris [45]
Hey there. So basically, find out how much the pencils and notebooks cost first.
The notebooks cost = $3.25
The pencils cost = $0.50

Then, think about what you need to figure out in this problem.
Jake has $20. You need to find how many notebooks Jake can buy in maximum after buying 8 pencils.

If Jake buys 8 pencils that costs $0.50 each, he spends $4 on the pencils.

So now, to find out how many notebooks he can buy, do 20 minus 4.
Jake's got $16 left.

If the notebooks cost $3.25 each, we need to find out how many notebooks he can buy by dividing them. So, 16 divided by 3.25 equals 4.923... and so on.

That means, Jake can buy 4 notebooks with his remaining money.
3 0
3 years ago
A video game randomly generates in-game items for players. Elizabeth records the items she receives. Based on the items Elisabet
Ilya [14]

Answer:

I have no idea what you think it is

7 0
2 years ago
Read 2 more answers
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
HELP ASAP!!!!! HELP FAST!!!!!
mr_godi [17]

Answer:

no they are not because 11, 14 is the only one that has a vertices

Step-by-step explanation:

8 0
3 years ago
Use for Questions 11-15
Juli2301 [7.4K]

Step-by-step explanation:

12. 136/300

=52%

13.45/300

=15%

6 0
2 years ago
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