Question

Determinar o valor de x​

Answer 1

Resposta:

2°)

a)

$\dfrac{2x - 3}{12} = \dfrac{8}{10} \\ \\( 2x -3 ) \times 10 = 8 \times 12 \\ \\ 20x - 30 =9 6 \\ \\ 20x = 96 - 30 \\ \\ 20x = 36 \\ \\ x = \frac{36 \div 4}{20 \div 4} \\ \\ \Large \boxed{ \green{x = \frac{9}{5} }} \\ \\ \frac{8}{10} = \frac{10}{y} \\ \\ 8y = 10 \times 10 \\ \\ 8y = 100 \\ \\ y = \frac{100 \div 4}{8 \div 4} \\ \\ \Large \boxed{ \green{y = \frac{25}{2} }}$

b)

$\dfrac{x}{10} = \dfrac{8}{5} \\ \\ 5x = 8 \times 10 \\ \\ 5x = 80 \\ \\ x = \frac{80}{5} \\ \\ \Large \boxed{ \green{ x = 16}} \\ \\ \dfrac{6}{y} = \dfrac{8}{5} \\ \\ 8y = 6 \times 5 \\ \\ 8y = 30 \\ \\ y = \frac{30 \div 2}{8 \div 2} \\ \\ \Large \boxed{ \green{ y = \frac{15}{4} }}$

3°)

a)

$\Large\bf Teorema \: de \: Pitágoras : \\\Large\boxed{ \bf {a}^{2} = {b}^{2} + {c}^{2} } \\ \\ {25}^{2} = {x}^{2} + {18}^{2} \\ \\ 625 = {x}^{2} + 324 \\ \\ 625 - 324 = {x}^{2} \\ \\ {x}^{2} = 625 - 324 \\ \\ {x}^{2} = 301 \\ \\ \Large \boxed{ \green{x = \sqrt{301} }}$

b)

${25}^{2} = {14}^{2} + {b}^{2} \\ \\ 625 = 196 + {b}^{2} \\ \\ 625 - 196 = {b}^{2} \\ \\ {b}^{2} = 429 \\ \\ \Large \boxed{ \green{ b = \sqrt{429} }}$

c)

${a}^{2} = {12}^{2} + {20}^{2} \\ \\ {a}^{2} = 144 + 400 \\ \\ {a}^{2} = 544 \\ \\ a = \sqrt{544} \\ a = \sqrt{16 \times 34} \\ \\ a = \sqrt{16} \times \sqrt{34} \\ \\ \Large \boxed{ \green{a = 4 \sqrt{34} }}$

4°)

$\dfrac{20 \div 4}{36 \div 4} = \\ \\ = \Large \boxed{ \green{ \dfrac{5}{9} }} \\ \\ \Large \boxed{ \underline{\blue{ \bf \: Bons \: Estudos!}}\:11/05/2021}$