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lesantik [10]
3 years ago
14

What is 13/25 in decimal form​

Mathematics
1 answer:
kenny6666 [7]3 years ago
8 0

Answer:

0.52

Step-by-step explanation:

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Select a counter-example that makes the conclusion false.. . 3 is prime, 5 is prime, 7 is prime. . Conclusion: All odd numbers a
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We must select a counter-example that makes the conclusion false. 

3 is a prime, 5 is a prime, and 7 is a prime. 
The conclusion is that all odd numbers are prime numbers. 
A counter example is that 9 is an odd number but it is NOT a prime number. 
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The answer is 266 but estimated is 260
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a certain stock starts the day at $27 3/8 per share. if it drops $2 1/2 during the day what is it’s closing value
Tasya [4]

let's firstly convert the mixed fractions to improper fractions and then simply get their difference, our denominators will be 8 and 2, so our LCD will be 8.

\bf \stackrel{mixed}{27\frac{3}{8}}\implies \cfrac{27\cdot 8+3}{8}\implies \stackrel{improper}{\cfrac{219}{8}}~\hfill \stackrel{mixed}{2\frac{1}{2}}\implies \cfrac{2\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{5}{2}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{219}{8}-\cfrac{5}{2}\implies \stackrel{\textit{using the LCD of 8}}{\cfrac{(1)219~~-~~(4)5}{8}}\implies \cfrac{219-20}{8}\implies \cfrac{199}{8}\implies 24\frac{7}{8}

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If y varies directly as x, and y is 400 when x is r and y is r when x is 4, what is the numeric constant of variation in this re
Kipish [7]
\bf \begin{array}{cccccclllll}
\textit{something}&&\textit{varies directly to}&&\textit{something else}\\ \quad \\
\textit{something}&=&{{ \textit{some value}}}&\cdot &\textit{something else}\\ \quad \\
y&=&{{ k}}&\cdot&x
\\
&&  y={{ k }}x
\end{array}\\\\
-----------------------------\\\\


\bf \textit{now, we also know }\qquad 
\begin{cases}
y=400\\
x=r\\
---\\
y=r\\
x=4
\end{cases}\implies 
\begin{cases}
400=kr\\
r=k4\\
----------\\
now\\
400=kr\implies \frac{400}{k}=\boxed{r}\\\\
thus\\\\
r=k4\implies \boxed{\frac{400}{k}}=4k
\end{cases}
\\\\\\
400=4k^2\implies \cfrac{400}{4}=k^2\implies 100=k^2\implies \sqrt{100}=k
\\\\\\
10=k\impliedby \textit{constant of variation}
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If one tree has the height that equals has, then the other one has the height of "h+20", which brings us to the sum = h + (h + 20)= 2h + 20
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