How does the value of a decimal changes as you move to the left or the right in a place value chart

1 answer:

7 0

If you move it to the left the number would increase so for example you had 2.3 and you move it to the left then it would be 23, so if you times it by tenth or hundredth then it would go to the left but if you divide then it would go down so moving to the right would get you 0.23

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Find the zeros of the polynomial function and state the multiplicity of each.

zheka24 [161]

$\bf f(x) = 4(x + 7)^2(x - 7)^3\implies 0 = 4(x + 7)^2(x - 7)^3
\\\\\\
0=(x+7)^2(x-7)^3\implies 
\begin{cases}
0=(x+7)^7\\
\qquad 0=(x+7)(x+7)\\
\qquad -7=x\\
\qquad -7=x\\
----------\\
0=(x-7)^3\\
\qquad 0=(x-7)(x-7)(x-7)\\
\qquad 7=x\\
\qquad 7=x\\
\qquad 7=x
\end{cases}$

the -7 is found there twice, so it has a multiplicity of 2
the +7 is there thrice, so it has multiplicity of 3

7 0

2 years ago

A rectangular prism has a volume of 16cm^3 and a total surface area of 40cm^2. A similar prism has length 10 in., width 5 in. an

krok68 [10]

The volume is twice that of the cube of width, so the width of the original prism is
.. ∛(16 cm³/2) = 2 cm

Length = 4 cm
Width = 2 cm
Height = 2 cm

5 0

3 years ago

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

Leya [2.2K]

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

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

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