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Serhud [2]
3 years ago
14

Which transformations of the graph of f(x)=-4^x result in the graph of f(x)=4^x+3?

Mathematics
2 answers:
GenaCL600 [577]3 years ago
7 0
The answer should be 1. Reflection over the x-axis and shifted 3 units to the left.
alex41 [277]3 years ago
3 0

Answer:

Option A is the answer

Step-by-step explanation:

Here f(x) =-4^x

in order to change the sign in front ,we need to  reflex it with respect to x axis

and since here 3 is added to x  [ as its x+3 in place of x ] which means  it change in x  which gives 3 units to the left.

  Option A is the answer.

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please Order least to greatest -2 3/10, -2 2/5, -2 -2 1/2, -3 And yes those are negative signs
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2 years ago
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A business was valued at £80000 at the start of 2013. In 5 years the value of this business raised to £95000. this is equivalent
Yuri [45]

the yearly increase of x% assumes is compounding yearly, so let's use that.

~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\dotfill &\£95000\\ P=\textit{original amount deposited}\dotfill &\£80000\\ r=rate\to r\%\to \frac{r}{100}\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{yearly, thus once} \end{array}\dotfill &1\\ t=years\dotfill &5 \end{cases}

95000=80000\left(1+\frac{~~ \frac{r}{100}~~}{1}\right)^{1\cdot 5}\implies \cfrac{95000}{80000}=\left( 1+\cfrac{r}{100} \right)^5 \\\\\\ \cfrac{19}{16}=\left( 1+\cfrac{r}{100} \right)^5\implies \sqrt[5]{\cfrac{19}{16}}=1+\cfrac{r}{100}\implies \sqrt[5]{\cfrac{19}{16}}=\cfrac{100+r}{100} \\\\\\ 100\sqrt[5]{\cfrac{19}{16}}=100+r\implies 100\sqrt[5]{\cfrac{19}{16}}-100=r\implies 3.5\approx r

4 0
2 years ago
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