Solve the equation X – 7 = 18 a. x = 11 b. x = -11 c. x = 25 d. x = -25

Mathematics

1 answer:

MakcuM [25]4 years ago

6 0

I believe the answer will be 'C', because 25-7 = 18.

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Leno4ka [110]

Answer:

(-0.0008)^0 =1
1291^1 =1
(1/3)^0 = 1

Step-by-step explanation:

$\mathrm{Apply\:exponent\:rule}:\quad \:a^0=1,\:\\\\(-0.0008)^0 = 1\\\\1291^0 =1\\\\-(-3141)^0 = -1\\\\-(\frac{5}{8} )^0 =-1\\\\0^1 = 0\\\\(\frac{1}{3} )^0 = 1\\$

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3 years ago

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atroni [7]

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3 years ago

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Drupady [299]

Mu=80, sigma=39, X=85
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8 0

3 years ago

A ribbon is cut into n parts. The length of each part increases such that they form a geometric progression. SPM Given the lengt

emmainna [20.7K]

Answer:

Step-by-step explanation:

$Let\ say\ r \ the \common\ ratio:\\a)a_4=r^2*a_2=9*a_2\Longrightarrow\ r=3\ (not\ -3\ since\ parts\ are\ positive)\\\\b)\\\displaystyle \sum_{i=1}^{n} a_i =a_1+a_2+a_3+...+a_n\\\\=a_1+a_1*r+a_1*r^2+a_1*r^3+...+a_1*r^{n-1}\\\\=a_1*(1+r+r^2+...+r^{n-1})\\\\=a_1*\dfrac{r^n-1}{r-1} \\\\147620=5*\frac{(3^n-1)}{3-1} \\\\3^n-1=\frac{2*147620}{5} \\\\3^n=59049 \\\\n*ln(3)=ln(59049)\\\\n=\dfrac{ln(59049)}{ln(3)} \\\\n=10\\\\a_n=5*3^{10-1}=5*19683=98415(cm)$

5 0

3 years ago

how much would $150 invested at 8% interest compounded anually be worth after 13 years? A(t)=P(1+r/n)^

PIT_PIT [208]

$\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 &\$150\\
r=rate\to 8\%\to \frac{8}{100}\to &0.08\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &13
\end{cases}
\\\\\\
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3 0

3 years ago