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Zora is budgeting her available time for the next couple days. She spent 2 over 5 of her time on homework and then 3 over 10 of her time on the internet. Homework and time on the internet consumed 14 hours. How much available time did Zora have to budget

x = budget

\(\begin{array}{rcll} \dfrac{2}{5}\cdot x + \dfrac{3}{10} \cdot x &=& 14\ h \\ x\cdot \left( \dfrac{2}{5} + \dfrac{3}{10} \right) &=& 14\ h \\ x\cdot \left( \dfrac{2\cdot 10+5\cdot 3}{5\cdot 10} \right) &=& 14\ h \\ x\cdot \left( \dfrac{20+15}{50} \right) &=& 14\ h \\ x\cdot \dfrac{35}{50} &=& 14\ h \\ x\cdot \dfrac{5\cdot 7}{5\cdot 10} &=& 14\ h \\ x\cdot \dfrac{ 7 }{ 10 } &=& 14\ h \quad & | \quad \cdot \dfrac{ 10 }{ 7 } \\ x &=& \dfrac{ 10 }{ 7 }\cdot 14\ h \\ x &=& 10\cdot 2\ h \\ \mathbf{ x }& \mathbf{=} & \mathbf{ 20\ h } \\ \end{array}\)

\(\begin{array}{rcll} 20\ h \cdot \dfrac{2}{5} &=& 8\ h \\\\ 20\ h \cdot \dfrac{3}{10} &=& 6\ h \\\\ 8\ h +6\ h &=&14\ h \end{array}\)

Zora has 20 hours

(10√1.3333-1)100=2.919 can someone explain to me how this happened please

\(\begin{array}{rcll} (\sqrt[10]{1.3333}-1)\cdot 100 &=& 100\cdot (\sqrt[10]{1.3333}-1) \\ &=& 100\cdot (1.02918343597-1) \\ &=& 100\cdot 0.02918343597 \\ &=& 2.918343597 \\ &\approx& 2.918 \end{array}\)

\(\begin{array}{rcll} (\sqrt[10]{1.3333\dots }-1)\cdot 100 &=& 100\cdot (\sqrt[10]{1.3333\dots}-1) \\ &=& 100\cdot (\sqrt[10]{1\frac{1}{3}}-1) \\ &=& 100\cdot (\sqrt[10]{\frac{4}{3}}-1) \\ &=& 100\cdot (1.02918600896-1) \\ &=& 100\cdot 0.02918600896 \\ &=& 2.91860089648 \\ &\approx& 2.919 \end{array}\)

Wie Rechnet man % wie Z.B 3% von 643  

\(\begin{array}{rcll} 3 \% \text{ von } 643 \\ \hline 3 \% \cdot 643 &=& 3\cdot \frac{1}{100} \cdot 643 \\ & = & \frac{3}{100} \cdot 643 \\ & = & \frac{3 \cdot 643 }{100} \\ & = & 3\cdot 6,43 \\ &=& 19,29 \end{array}\)

Make the number 48 with the numbers 1, 6, 7, 8.

Only use addition, subtraction, multiplication, and division. You can use each number once! You have to use all the numbers

\(\begin{array}{rcll} 48 = \dfrac{8} { \left( \dfrac{7}{6} \right) - 1 } \\ \end{array}\)

\(\begin{array}{rcll} 48 = \dfrac{6} { 1- \left( \dfrac{7}{8} \right) } \\ \end{array}\)

Make the number 48 with the numbers 1, 6, 7, 8.
Only use addition, subtraction, multiplication, and division. You can use each number once!

How do you determine the correct way to simplify two radical numbers while dividing them and how do I know if the number remains squared or becomes a whole number? Perhaps this should be used for an example? 4√63 / 6√36

\(\begin{array}{rcll} \dfrac{ 4 \cdot \sqrt{63} } { 6 \cdot \sqrt{36} } &=& \dfrac{ 4 }{ 6 }\cdot \dfrac{ \sqrt{63} } { \sqrt{36} } \\\\ &=& \dfrac{ 2 }{ 3 }\cdot \dfrac{ \sqrt{63} } { \sqrt{36} } \\\\ &=& \dfrac{ 2 }{ 3 }\cdot \sqrt { \dfrac{ 63 } { 36 } } \\\\ &=& \dfrac{ 2 }{ 3 }\cdot \sqrt { \dfrac{ 9\cdot 7 } { 9\cdot 4 } } \\\\ &=& \dfrac{ 2 }{ 3 }\cdot \sqrt { \dfrac{ 7 } { 4 } } \\\\ &=& \dfrac{ 2 }{ 3 }\cdot \dfrac{ \sqrt{7} } { \sqrt{4} } \\\\ &=& \dfrac{ 2 }{ 3 }\cdot \dfrac{ \sqrt{7} } { 2 } \\\\ &=& \dfrac{ \sqrt{7} } { 3 } \\\\ \end{array}\)

Make the number 48 with the numbers 1, 6, 7, 8.

Only use addition, subtraction, multiplication, and division.

You can use each number once!

Make the number 48 with the numbers 1, 6, 7, 8.

Only use addition, subtraction, multiplication, and division.

You can use each number once!

You have to use all the numbers

\(\begin{array}{rcll} 48 = \dfrac{6} { 1- \left( \dfrac{7}{8} \right) } \\ \end{array}\)

x+2y=7

2x+2y=13

\(\begin{array}{lrcll} (1)& x+2y &=& 7 \quad & | \quad -x\\ & 2y &=& 7 -x \\\\ (2)& 2x+2y &=& 13 \\ & 2x +(7-x) &=& 13 \\ & 2x + 7-x &=& 13 \\ & x + 7 &=& 13 \quad & | \quad -7\\ & x &=& 13 -7\\ & \mathbf{x} & \mathbf{=} & \mathbf{6} \\\\ & 2y &=& 7 -x \\ & 2y &=& 7 -6 \\ & 2y &=& 1 \quad & | \quad :2\\ & y &=& \frac12\\ & \mathbf{y} & \mathbf{=} & \mathbf{0.5} \\\\ \end{array}\)

a cookie requires 1\2 cup of sugar. whitch is equivalent to 1\2

2\3?

3\5?

4\8? or

6\10

\(\dfrac{1}{2} \cdot \dfrac{4}{4} = \dfrac{1\cdot 4}{2\cdot 4}\\\\ \mathbf{=\dfrac{4}{8}}\)