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3 years ago

Which of the following can be used to find the absolute value of 4 - 7i?

Mathematics

2 answers:

fomenos3 years ago

7 0

Answer:

Option 1 is correct

Explanation:

We have been given with the function 4 -7i

To find the absolute value means to find the modulus of the function.

And modulus of the function is $\sqrt{a^2+b^2}$

Here we have a=4 and b= -7 so substituting the values in the formula we will get

$\sqrt{4^2+(-7)^2}$

Hence, Option 1 is correct

Option 2 is incorrect because we do not consider "i" in finding absolute value or modulus

Option 3 is incorrect because it should be -7 not 7

Option 4 is incorrect because it should not be whole square.

Therefore Option 1 is correct.

inn [45]3 years ago

5 0

The first one.

$|a+bi|=\sqrt{a^2+b^2}$

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3 years ago

Solve the following equation by completing the square. 3x^2-3x-5=13

mr Goodwill [35]

we'll start off by grouping some

$\bf 3x^2-3x-5=13\implies (3x^2-3x)-5=13\implies 3(x^2-x)-5=13 \\\\\\ 3(x^2-x)=18\implies (x^2-x)=\cfrac{18}{3}\implies (x^2-x)=6\implies (x^2-x+~?^2)=6$

so we have a missing guy at the end in order to get the a perfect square trinomial from that group, hmmm, what is it anyway?

well, let's recall that a perfect square trinomial is

$\bf \qquad \textit{perfect square trinomial} \\\\ (a\pm b)^2\implies a^2\pm \stackrel{\stackrel{\text{\small 2}\cdot \sqrt{\textit{\small a}^2}\cdot \sqrt{\textit{\small b}^2}}{\downarrow }}{2ab} + b^2$

so we know that the middle term in the trinomial, is really 2 times the other two without the exponent, well, in our case, the middle term is just "x", well is really -x, but we'll add the minus later, we only use the positive coefficient and variable, so we'll use "x" to find the last term.

$\bf \stackrel{\textit{middle term}}{2(x)(?)}=\stackrel{\textit{middle term}}{x}\implies ?=\cfrac{x}{2x}\implies ?=\cfrac{1}{2}$

so, there's our fellow, however, let's recall that all we're doing is borrowing from our very good friend Mr Zero, 0, so if we add (1/2)², we also have to subtract (1/2)²

$\bf \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2-\left[ \cfrac{1}{2} \right]^2 \right)=6\implies \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2 \right)-\left[ \cfrac{1}{2} \right]^2=6 \\\\\\ \left(x-\cfrac{1}{2} \right)^2=6+\cfrac{1}{4}\implies \left(x-\cfrac{1}{2} \right)^2=\cfrac{25}{4}\implies x-\cfrac{1}{2}=\sqrt{\cfrac{25}{4}} \\\\\\ x-\cfrac{1}{2}=\cfrac{\sqrt{25}}{\sqrt{4}}\implies x-\cfrac{1}{2}=\cfrac{5}{2}\implies x=\cfrac{5}{2}+\cfrac{1}{2}\implies x=\cfrac{6}{2}\implies \boxed{x=3}$

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3 years ago