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aleksandr82 [10.1K]
3 years ago
11

Can some one help me

Mathematics
1 answer:
lara [203]3 years ago
5 0

Answer:

there nothing to solve

Step-by-step explanation:

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I need help solving it
blagie [28]
X2-64=0
x2=64

x=8 because when multiplied by itself, and plugged back in, it works
8 0
3 years ago
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Please answer this ASAP
Amanda [17]

Answer: 2x and y= 4+

Step-by-step explanation:

x = 2 times 2 and so on....

y = 4 plus 4 plus 4 etc....

3 0
3 years ago
Which of the following expressions does not equal: the sixth root of 81x^4y^8
DerKrebs [107]
\sqrt[6]{81x^4y^8}=\left[3^4x^4(y^2)^4\right]^\frac{1}{6}=\left[\left(3xy^2\right)^4\right]^\frac{1}{6}\\\\=\left(3xy^2\right)^{4\cdot\frac{1}{6}}=\left(3xy^2\right)^\frac{2}{3}\to A\\\\=\left(3xy^2\right)^\frac{2}{3}=\sqrt[3]{\left(3xy^2\right)^2}=\sqrt[3]{9x^2y^4}\to D\\\\=\left(3xy^2\right)^\frac{2}{3}=\left(3x\right)^\frac{2}{3}y^{2\cdot\frac{2}{3}}=\left(3x\right)^\frac{2}{3}y^\frac{4}{3}\to B\\\\\\Answer:C
4 0
4 years ago
Match the pairs of polynomials to their products. (xy + 9y + 2) and (xy – 3) x2y2 + 3x2y – 7xy – 27x – 18 (2xy + x + y) and (3xy
zlopas [31]

Answer:

Step-by-step explanation:

1) (xy+ 9y + 2) and (xy – 3)

Each term of second expression will be multiplied by first bracket.

xy(xy+9y+2) -3(xy+9y+2)

x²y²+9xy²+2xy-3xy-27y-6

x²y²+9xy²-xy-27y-6

2) (2xy + x + y) and (3xy2 – y)

3xy²(2xy+x+y) -y(2xy+x+y)

6x²y³+3x²y²+3xy³-2xy²-xy-y²

6x²y³ – 2xy² + 3x²y² – xy + 3xy³ – y²

3)  (x – y) and (x + 3y)

x(x-y) +3y(x-y)

x²-xy+3xy-3y²

x²+2xy-3y²

4) (xy + 3x + 2) and (xy – 9)

xy(xy + 3x + 2) -9(xy + 3x + 2)

x²y²+3x²y+2xy-9xy-27x-18

x²y²+3x²y-7xy-27x-18

5) (x2 + 3xy – 2) and (xy + 3)

xy(x2 + 3xy – 2) +3(x2 + 3xy – 2)

x³y+3x²y²-2xy+3x²+9xy-6

x³y+3x²+3x²y²+7xy-6

6) (x + 3y) and (x – 3y)

x(x + 3y) -3y(x + 3y)

x²+3xy-3xy-9y²

x²-9y² ....

6 0
4 years ago
At the beginning of each year for 14 years, Sherry Kardell invested $400 that earns 10% annually. What is the future value of Sh
Naddik [55]
Hmm if I'm not mistaken, is just an "ordinary" annuity, thus 

\bf \qquad \qquad \textit{Future Value of an ordinary annuity}
\\\\
A=pymnt\left[ \cfrac{\left( 1+\frac{r}{n} \right)^{nt}-1}{\frac{r}{n}} \right]
\\\\\\

\bf \begin{cases}
A=
\begin{array}{llll}
\textit{original amount}\\
\textit{already compounded}
\end{array} &
\begin{array}{llll}

\end{array}\\
pymnt=\textit{periodic payments}\to &400\\
r=rate\to 10\%\to \frac{10}{100}\to &0.1\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, so once}
\end{array}\to &1\\

t=years\to &14
\end{cases}
\\\\\\
A=400\left[ \cfrac{\left( 1+\frac{0.1}{1} \right)^{1\cdot 14}-1}{\frac{0.1}{1}} \right]

3 0
3 years ago
Read 2 more answers
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