What is 0.9×10 to the negative power of 1 in standard notation

2 answers:

5 0

$a^{-1}=\dfrac{1}{a}\\\\0.9\cdot10^{-1}=0.9\cdot\dfrac{1}{10}=\dfrac{0.9}{10}=\dfrac{0.9\cdot10}{10\cdot10}=\dfrac{9}{100}=0.09$

andreev551 [17]3 years ago

3 0

0.09 is the answer …

You might be interested in

Let v1 =(-6,4) and v2=(-3,6) compute the following what i sthe angle between v1 and v2

Agata [3.3K]

$\bf ~~~~~~~~~~~~\textit{angle between two vectors }
\\\\
cos(\theta)=\cfrac{\stackrel{\textit{dot product}}{u \cdot v}}{\stackrel{\textit{magnitude product}}{||u||~||v||}} \implies 
\measuredangle \theta = cos^{-1}\left(\cfrac{u \cdot v}{||u||~||v||}\right)\\\\
-------------------------------$

$\bf \begin{cases}
v1=\ \textless \ -6,4\ \textgreater \ \\
v2=\ \textless \ -3,6\ \textgreater \ \\
------------\\
v1\cdot v2=(-6\cdot -3)+(4\cdot 6)\\
\qquad \qquad 42\\
||v1||=\sqrt{(-6)^2+4^2}\\
\qquad \sqrt{52}\\
||v2||=\sqrt{(-3)^2+6^2}\\
\qquad \sqrt{45}
\end{cases}\implies \measuredangle \theta =cos^{-1}\left( \cfrac{42}{\sqrt{52}\cdot \sqrt{45}} \right)
\\\\\\
\measuredangle \theta =cos^{-1}\left( \cfrac{42}{\sqrt{2340}} \right)\implies \measuredangle \theta \approx 29.74488129694^o$

8 0

3 years ago

Find the missing number:

nirvana33 [79]

Answer:

20%

Step-by-step explanation:

Not sure if this is exactly correct, but let's say we start with 1 unit. If it increased by 25%, we get:

1 × 1.25 = 1.25 units.

Now, to get back to 1 unit, we need to find:

1.25 × ? = 1

? = $\frac{1}{1.25}$ = 0.80