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Mrac [35]
2 years ago
13

Given the expression: 6x10 − 96x2

Mathematics
1 answer:
Anettt [7]2 years ago
5 0

Step-by-step explanation:

Part A:

I'm going to assume you meant to write 6x^10 - 96x^2. Based off that assumption you can factor out 6 from both equations as well as x^2.

This is going to give you

6x^2(x^8-16). I

Part B:

if you look at the inside, both of the values are perfect squares (if the exponent is even, it's a perfect square). So the equation can further be rewritten using the difference of squares

6x^2(x^4-4)(x^4+4)

If you look at the expression, you'll notice one of the equations can further be rewritten as the difference of squares (x^4 - 4).

6x^2(x^2-2)(x^2+2)(x^4+4)

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Answer:

Read this section of the text:

These raptors eat primarily flying insects, so they do most of their hunting on the wing.

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Step-by-step explanation:

4 0
3 years ago
dude, I just got almost all of my points taken away from somebody that has 0 answers, 0 questions, and almost 2 million points..
Finger [1]

Answer:

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Step-by-step explanation:

7 0
3 years ago
PLSSSSSS HELPPPP......Which equation can be used to find the unknown length, b, in this triangle?
castortr0y [4]
Its A. the side lengths squared equal the hypotenuse squared.
6 0
3 years ago
Read 2 more answers
Suppose that you have $6000 to invest. Which investment yields the greater return over four years: 8.25% compounded quarterly or
WARRIOR [948]
\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$6000\\
r=rate\to 8.25\%\to \frac{8.25}{100}\to &0.0825\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{quarterly, thus four}
\end{array}\to &4\\
t=years\to &4
\end{cases}
\\\\\\
A=6000\left(1+\frac{0.0825}{4}\right)^{4\cdot 4}\implies A=6000(1.020625)^{16}\\\\
-------------------------------\\\\



\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$6000\\
r=rate\to 8.3\%\to \frac{8.3}{100}\to &0.083\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{semiannually, thus two}
\end{array}\to &2\\
t=years\to &4
\end{cases}
\\\\\\
A=6000\left(1+\frac{0.083}{2}\right)^{2\cdot 4}\implies A=6000(1.0415)^8

compare them away.
4 0
3 years ago
a country population in 1993 was 94 million in 1999 it was 99 million estimate the population in 2005 using the exponential grow
tester [92]
K = ln(ending amt/bgng amount)/6 years
k = ln(99,000,000/94,000,000) / 6 years
k = ln( <span> <span> <span> 1.0531914894 </span> </span> </span> ) / 6 years
k = <span> <span> <span> 0.0518250679 </span> </span> </span> / 6
k = <span> <span> <span> 0.0086375113</span></span></span>

Final population = bgng*e^k*years
Where "e" is the mathematical constant 2.718281828
Final population = 94,000,000*e^<span>0.0086375113*12</span>
Final population = 94,000,000*e^<span><span><span>0.1036501356 </span> </span> </span>
Final population = 94,000,000*  <span> <span> <span> 1.1092123132 </span> </span> </span> <span><span> </span> </span>
Final population = <span> <span> <span> 104,265,957 </span> </span> </span>


5 0
3 years ago
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