What the answer plzzhelppp

Mathematics

1 answer:

AfilCa [17]2 years ago

4 0

Step-by-step explanation:

You have to find the decimals of them

So the 3rd on is correct

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Write a rule for a geometric sequence that has -6 as its first term and a common ratio of 2. What is the tenth term of this sequ

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The general form of the geometric sequence is $a_{n} = a_{1} (r) ^{n-1}$ , where a sub n is the number term you're looking for (we're looking for the tenth term). a sub 1 is the first term in the sequence (ours is -6), r is the common ratio, and n-1 is the numbered term you're looking for minus 1. Our formula then looks like this: $a_{10} =-6(2) ^{10-1}$ . Simplify it to $a_{10} =-6(2)^9$ . Take 2 to the 9th power then multiply it by -6 to get -3072. C is your answer.

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Prove that √2 +√5 is irrational

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We have to prove that $\sqrt{2}+\sqrt{5}$ is irrational. We can prove this statement by contradiction.

Let us assume that $\sqrt{2}+\sqrt{5}$ is a rational number. Therefore, we can express:

$a=\sqrt{2}+\sqrt{5}$

Let us represent this equation as:

$a-\sqrt{2}=\sqrt{5}$

Upon squaring both the sides:

$(a-\sqrt{2})^{2}=(\sqrt{5})^{2}\\a^{2}+2-2\sqrt{2}a=5\\a^{2}-2\sqrt{2}a=3\\\sqrt{2}=\frac{a^{2}-3}{2a}$

Since a has been assumed to be rational, therefore, $\frac{a^{2}-3}{2a}$ must as well be rational.

But we know that $\sqrt{2}$ is irrational, therefore, from equation $\sqrt{2}=\frac{a^{2}-3}{2a}$ the expression $\frac{a^{2}-3}{2a}$ must be irrational, which contradicts with our claim.