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Harrizon [31]
3 years ago
15

You are given the equation \sqrt{76+n } and \sqrt{2n + 26}. What is the smallest value of n that will make each number rational?

30 points!
Mathematics
1 answer:
Diano4ka-milaya [45]3 years ago
4 0

Answer:

A proof that square root of 2 is irrational. ... An equation x² = a, and the principal square root ... The number under the radical sign is called the radicand. ... the given square numbers, each product of square numbers is equal to what square ... are relatively prime -- and it will be impossible to divide n· n into m· m and get 2.

Step-by-step explanation:

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I need help with the first question
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Examine the diagram.
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Answer:

m\angle 1 =60.75^\circ \\m\angle 2 =119.25^\circ

Step-by-step explanation:

From the diagram:

m\angle 1 =60.75^\circ $(Corresponding Angles)\\

Now Angle 1 and Angle 2 are on a straight line, therefore:

m\angle 1 +m\angle 2 =180^\circ $(Linear Postulate)\\Since m\angle 1=60.75^\circ\\60.75^\circ +m\angle 2 =180^\circ\\m\angle 2=180^\circ-60.75^\circ\\m\angle 2 =119.25^\circ

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Read 2 more answers
Find all solutions of the equation in the interval [0,2π) sec(theta) +2=0
Komok [63]
I will first reveal the most obvious secret of trig: Most of the problems are 30/60/90 or 45/45/90 triangles of some sort.

\sec \theta + 2  = 0

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The rule to remember is \cos x = \cos a has solutions

x = \pm a + 2 \pi k \quad integer k

Continuing where we left off, a cosine of -1/2 is a 30/60/90 triangle in the second or third quadrant; we pick second, and a little thought tells us the angle is  120^\circ.

\cos \theta = - \frac 1 2

\cos \theta = \cos \frac {2\pi} 3

\theta = \pm \frac {2\pi} 3 + 2 \pi k \quad integer k

In the range we want, that's k=0 with the plus sign and k=1 with the minus, so 

\theta = \frac {2\pi} 3 or -\frac {2\pi} 3 + 2 \pi = \frac{4 \pi} 3

\theta = \frac {2\pi} 3 or \frac{4 \pi} 3
6 0
3 years ago
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