Question

Ayudenme por favor
∛3435 - √100.2=
√49 . 2² + (3.5)²=
√25 - 4 + 3 . (2²+3²)=

Answer 1

Ayudenme por favor

∛3435 - √100.2=

√49 . 2² + (3.5)²=

√25 - 4 + 3 . (2²+3²)=

Hola!!

Vamos a resolver paso a paso tus ejercicios

$\sqrt[3]{3435} -\sqrt{100} . 2 = \qquad la \ raiz \ cubica\ de \ 3435 =NO \ es\ entera, entonces\\\\\\ \sqrt[3]{3435} -\sqrt{100} . 2 = 15,08 - 10. 2 \to resolvemos \ las \ raices \\\\ \sqrt[3]{3435} -\sqrt{100} . 2 = 15,08 - 20 \to resolvemos \ la \ multiplicacion \\\\\boxed{ \sqrt[3]{3435} -\sqrt{100} . 2 = -4,92}\to resolvemos \ la \ resta .$

$\sqrt{49}. 2^2 + (3.5)^2= \to resolvemos \ la \ raiz \ y \ las\ potencias\\\\\sqrt{49}. 2^2 + (3.5)^2= 7 . 4 + (3^2.5^2) = \ resolvemos \ las \ potencias\\\\ \sqrt{49}. 2^2 + (3.5)^2= 7.4 + (9.25) = \to resolvemos \ las \ multiplicaciones\\\\ \sqrt{49}. 2^2 + (3.5)^2= 28 + 225 \to resolvemos \ la \ suma \\\\ \boxed{\sqrt{49}. 2^2 + (3.5)^2=253 }$

$\sqrt{25}- 4 +3.(2^2+3^2)= \to resolvemos \ la \ raiz \ y \ las\ potencias\\\\\sqrt{25}- 4 +3.(2^2+3^2)= 5-4+3. (4+9) \to resolvemos \ los \ parentesis \\\\ \sqrt{25}- 4 +3.(2^2+3^2)= 5-4+3. (13) \to resolvemos \ la \ multiplicacion\\\\ \sqrt{25}- 4 +3.(2^2+3^2)= 5-4+39\\\\ \boxed{\sqrt{25}-4+3.(2^2+3^2)=40 }$

Espero que te sirva, salu2!!!!