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2 years ago

S+f=24 s=3f I need this now

Mathematics

1 answer:

scoundrel [369]2 years ago

4 0

Answer:

f=6 and s=18

Step-by-step explanation:

Let's solve your system by substitution.

s+f=24;s=3f

Rewrite equations:

s=3f;f+s=24

Step: Solves=3ffor s:

s=3f

Step: Substitute3fforsinf+s=24:

f+s=24

f+3f=24

4f=24(Simplify both sides of the equation)

(Divide both sides by 4)

f=6

Step: Substitute6forfins=3f:

s=3f

s=(3)(6)

s=18(Simplify both sides of the equation)

Answer:

f=6 and s=18

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$\huge \boxed{\mathfrak{Question} \downarrow}$

Expand & simplify ⇨ $( \sqrt{10} - \sqrt{2} ) ^{2}$ . Give your answer in the form $b - c \: \sqrt{5}$ where b & c are integers.

$\large \boxed{\mathbb{ANSWER\: WITH\: EXPLANATION} \downarrow}$

$( \sqrt { 10 } - \sqrt { 2 } ) ^ { 2 }$

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(\sqrt{10}-\sqrt{2}\right)^{2}$ .

$\left(\sqrt{10}\right)^{2}-2\sqrt{10}\sqrt{2}+\left(\sqrt{2}\right)^{2}$

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Factor $10=2\times 5$ . Rewrite the square root of the product $\sqrt{2\times 5}$ as the product of square roots $\sqrt{2}\sqrt{5}$ .

$10-2\sqrt{2}\sqrt{5}\sqrt{2}+\left(\sqrt{2}\right)^{2}$

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$\boxed{ \boxed{\bf\:12-4\sqrt{5} }}$

Here, b & c are integers where $\boxed{ \sf \: b = 12 \: and \: c = 4}$

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