The surface area is a sum of areas of four triangles with base of <em>8 m</em> and hight of <em>12 m</em> and areas of five squares with sides of <em>8 m.</em>

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6 0

2 years ago

A Venn diagram is shown below: What are the elements of (A n B) ‘ ?

vesna_86 [32]

The elements of (A n B)' are ( 3, 4 , 5 ,6). Option A

<h3>How to determine the set</h3>

The elements of this set (A n B) explains the common elements of both sets without repetition

Set A = 3, 4

Set B = 5, 6

A n B = 1, 2

(A n B)' = Is the elements both A and B in common but is not found in the universal set

(A n B)' = ( 3, 4 , 5 ,6)

Thus, the elements of (A n B)' are ( 3, 4 , 5 ,6). Option A

Learn more about Venn diagrams here:

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8 0

1 year ago

Prove that √2 +√5 is irrational

Sindrei [870]

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.

Therefore, by contradiction, $\sqrt{2}+\sqrt{5}$ is irrational.

4 0

2 years ago