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4 years ago

7

Prove that root 7 is an irrational number.

2 answers:

HACTEHA [7]4 years ago

8 0

Answer:

it is an irrational number and here is why

Step-by-step explanation:

So think of square root as it's answer is something that you you multiply its self by its self and you get that answer. Ex: 8 x 8 = 64 square root of 64 is 8. So in that case there is nothing you can exactly multiply by itself to get 7 so in that case it's a decimal or an irrational number.

HOPE THIS HELPS!

s344n2d4d5 [400]4 years ago

4 0

Answer: Root 7 would equal 2.64575131106 and would go on forever.

Step-by-step explanation: It isn't one of the perfect squares. For example, the first 10 perfect square are 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100.

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Algebra 2 help. Solving equations containing two radicals

Marysya12 [62]

These are 3 questions and 3 answers.

Question 1.

Answer: bot equations have the same potential solutions, but equation A may have extraneous solutions

Explanation:

1) Equation A

$\sqrt{x^2+3x-6}=\sqrt{x+2}\\ \\ x^2+3x-6=x+2\\ \\ x^2+2x-8=0\\ \\ (x+4)(x-2)=0\\ \\ x=-4 ; x =2$

Replace x = -4 in the right side of the equation:

$\sqrt{x+2} =\sqrt{-4+2} =\sqrt{-2}$

Which is not defined, so this is a extraneous solution.

2) Equation B.

Since it is a cube root, it is defined for any (negative, zero, and positive) values of x.

Question 2.

Answer: cubing both sides once.

Explanation:

$\sqrt[3]{5x-2} -\sqrt[3]{4x} =0\\ \\ \sqrt[3]{5x-2} =\sqrt[3]{4x}\\ \\ (\sqrt[3]{5x-2} )^3=(\sqrt[3]{4x})^3\\ \\ 5x-2=4x\\ \\ 5x-4x=2\\ \\ x=2$

3) Question 3.

Answer: c - 2 = 25 + c + 10√c

Explanation:

$\sqrt{c-2} -\sqrt{c} =5\\ \\ \\ \sqrt{c-2} =\sqrt{c} +5\\ \\ (\sqrt{c-2})^{2} =(\sqrt{c}+5)^2\\ \\ c-2=c+10\sqrt{c} +25\\ \\ c-2=25+c+10\sqrt{c}$

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3 years ago

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siniylev [52]

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nekit [7.7K]

Answer:

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Step-by-step explanation:

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fredd [130]

Answer:

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Step-by-step explanation:

250x1.3687=342.175

342.175-150=192.175

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