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castortr0y [4]
3 years ago
9

Write the ratio as a ratio of a whole numbers using fractional notation 6 1/2 to 15 1/6

Mathematics
1 answer:
motikmotik3 years ago
5 0
\bf 6\frac{1}{2}:15\frac{1}{6}\qquad 
\begin{cases}
6\frac{1}{2}\implies \cfrac{6\cdot 2+1}{2}\\\\
15\frac{1}{6}\implies \cfrac{15\cdot 6+1}{6}
\end{cases}\implies \cfrac{ \frac{6\cdot 2+1}{2}}{\frac{15\cdot 6+1}{6}}\\\\
-----------------------------\\\\
\cfrac{\frac{a}{b}}{\frac{c}{{{ d}}}}\implies \cfrac{a}{b}\cdot \cfrac{{{ d}}}{c}\qquad thus
\\\\\\
\cfrac{6\cdot 2+1}{2}\cdot \cfrac{6}{15\cdot 6+1}

simplify away
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2 years ago
Which of the following is equal to the expression above
EleoNora [17]

Answer:

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Step-by-step explanation:

{(27 \times 250)}^{ \frac{1}{3} }  =  {(27 \times 125 \times 2)}^{ \frac{1}{3} }  \\  =  {27}^{ \frac{1}{3} }  \times  {125}^{ \frac{1}{3} }  \times  {2}^{ \frac{1}{3} }  \\  =  \sqrt[ 3]{27}  \times  \sqrt[3]{125}  \times  \sqrt[3]{2}  \\  =  \sqrt[3]{ {3}^{3} }  \times  \sqrt[3]{ {5}^{3} }  \times  \sqrt[3]{2}  \\  = 3 \times 5 \times  \sqrt[3]{2}  \\  = 15 \sqrt[3]{2}

4 0
3 years ago
If the initial amount of iodine-131 is 537 grams , how much is left after 10 days?
viktelen [127]

Answer:

225.78 grams

Step-by-step explanation:

To solve this question, we would be using the formula

P(t) = Po × 2^t/n

Where P(t) = Remaining amount after r hours

Po = Initial amount

t = Time

In the question,

Where P(t) = Remaining amount after r hours = unknown

Po = Initial amount = 537

t = Time = 10 days

P(t) = 537 × 2^(10/)

P(t) = 225.78 grams

Therefore, the amount of iodine-131 left after 10 days = 225.78 grams

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