A triangle has a side length of 3/4 inch and a side length of 3 inches. What could be the length, in inches, of the third side?

1 answer:

6 0

To solve this problem you must apply the proccedure shown below:

1- If the triangle is a rigth triangle, you must apply the Pythagorean Theorem. Taking the length 3 inches as the hypotenuse, which is the longest side of the triangle, you have:

a^2=b^2+c^2
c=√(a^2-b^2)

b=3/4 inches
a=3 inches

2- When you susbtitute the values, you obtain:

c=√(3 inches)^2-(3/4 inches)^2
c=2.90 inches

The answer is: 2.90 inches.

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Dmitriy789 [7]

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Step-by-step explanation:

Ugh! Wow, this is going to be tedious, thanks alot, bro (jk, I got your back).

x - 99 ≤ -104

+ 99 + 99

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There should be a line going from -5 to negative infinity (AKA the left) with a FILLED circle. So, the second one is correct.

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There should be an arrow with a FILLED circle going to negative infinity (AKA the left). So, the first one is correct.

I'm going to take a shortcut and notice that one of the lines has a filled circle while the other one has an empty circle. So the empty circle must relate to the question without a ≤ or ≥, but with a < or >. We see that 75 < 69 doesn't have ≤ or ≥ but a '<.' So this one must have the empty circle, which is on line 3. The last equation has to be on line 4.

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Answer:

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Step-by-step explanation:

$3x^2-2x-1=0\\\\3(x^2-\frac{2}{3}x-\frac{1}{3})=0\\ \\3(x^2-\frac{2}{3}x-\frac{1}{3}+\frac{4}{9})=3(\frac{4}{9})\\ \\ 3(x^2-\frac{2}{3}x+\frac{1}{9})=\frac{4}{3}\\ \\ 3(x-\frac{1}{3})^2=\frac{4}{3}\\ \\(x-\frac{1}{3})^2=\frac{4}{9}\\ \\x-\frac{1}{3}=\pm\frac{2}{3}$