How can 1/2 be written in decimal

Mathematics

2 answers:

harina [27]4 years ago

7 0

Answer:

Step-by-step explanation:

.5 is the answer because decimals have digits counting by 10's.....if that makes sense. Like whole numbers start with ones then tens, hundereds, thousands and so on. Decimals go 10ths, 100ths, and that goes on. So 1/2 of 10 is 5 another example is .10 in half (half is the same thing as dividing by two.) so you get .5

Im not the best at explaining things lol, but I hope this helps and havea great day!

Burka [1]4 years ago

3 0

Answer:

Step-by-step explanation:

We need to get the denominator to be a power of 10

1/2 *5/5 = 5/10

5/10 = .5

1/2 = .5

Another approach is to take the numerator and divide by the denominator.

-----------------

2 | 1.0000

2 goes into 1 0 times

-----------------

2 | 1.0000

2 goes into 10 5 times

0.5

-----------------

2 | 1.0000

--------------

And there is 0 left

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1/3 radical 176xy to the 8th power

Tanya [424]

Answer:

Remove all perfect squares from inside the square root. ... I think it's about eighth or ninth grade. ... so if you have the cube root of the square root of (x-5) =2, you get ((x-5)^(1/2))^1/3 = 2, power to power requires multiplication, so (x-5)^1/6 = 2, ...

Missing: 176 ‎xy

5 0

3 years ago

How does the graph of f(x)=3lx+2l+4 relate to its parent function?

Sergeeva-Olga [200]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\

\begin{array}{rllll} 
% left side templates
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}
\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\end{array}$

$\bf \begin{array}{llll}


\bullet \textit{ vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}
\end{array}$

now, with that template above in mind, let's see this one

$\bf parent\implies f(x)=|x|
\\\\\\
\begin{array}{lllcclll}
f(x)=&3|&1x&+2|&+4\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}$

A=3, B=1, shrunk by AB or 3 units, about 1/3
C=2, horizontal shift by C/B or 2/1 or just 2, to the left
D=4, vertical shift upwards of 4 units

check the picture below