Please help ASAP!!!!!!!!!!

Mathematics

1 answer:

zheka24 [161]3 years ago

5 0

Answer:

D. Angle P, Angle OPQ, Angle QPO

Step-by-step explanation:

The angle is at P so P must be either between the two other letters or be by itself so D is the correct answer.

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I think 4.2 x 10^6 is the answer you're looking for.

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3 years ago

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Suppose Kaitlin places $6500 in an account that pays 12% interest compounded each year.

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$\bf ~~~~~~ \textit{Compound Interest Earned Amount \underline{for 1 year}} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$6500\\ r=rate\to 12\%\to \frac{12}{100}\dotfill &0.12\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &1 \end{cases}$

$\bf A=6500\left(1+\frac{0.12}{1}\right)^{1\cdot 1}\implies A=6500(1.12)\implies A=7280 \\\\[-0.35em] ~\dotfill$

$\bf ~~~~~~ \textit{Compound Interest Earned Amount \underline{for 2 years}} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$6500\\ r=rate\to 12\%\to \frac{12}{100}\dotfill &0.12\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &2 \end{cases}$

$\bf A=6500\left(1+\frac{0.12}{1}\right)^{1\cdot 2}\implies A=6500(1.12)^2\implies A=8153.6$

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For part a: you just need to find how far the vertex has been moved from the origin, or the point (0,0). As the vertex is at the point (2,-3), it has been translated right 2 horizontally and down 3 vertically.

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Now that you have h and k, substitute them back into the equation.
Your final answer for part b is: y=(x-2)^2 -3.

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3 years ago

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Fudgin [204]

D would be the answer.

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