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Luda [366]
3 years ago
13

Anyone know this ? picture shown please help

Mathematics
1 answer:
valina [46]3 years ago
5 0

They're just after some straightforward algebra starting from the Law of Cosine.  We don't have to evaluate any trig functions or anything like that.

c^2 = a^2 + b^2 - 2ab \cos C

We move the c² and the cosine term to their other sides,

2ab \cos C = a^2 + b^2- c^2

That's it for the algebra.  

a,b and c are the side lengths opposite A,B and C respectively.  So a=4, b=5, c=2.

We substitute:

2ab \cos C = 4^2 + 5^2- 2^2 = 16 + 25 - 4 = 37

Answer: A. 37

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A jar contains 8 marbles marked with the numbers 1 through 8 . You pick a marble at random. What is the probability of not picki
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7/8 Because there are (7 marbles that are not 8 / 8 possibilities).
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If the sum of the zereos of the quadratic polynomial is 3x^2-(3k-2)x-(k-6) is equal to the product of the zereos, then find k?
lys-0071 [83]

Answer:

2

Step-by-step explanation:

So I'm going to use vieta's formula.

Let u and v the zeros of the given quadratic in ax^2+bx+c form.

By vieta's formula:

1) u+v=-b/a

2) uv=c/a

We are also given not by the formula but by this problem:

3) u+v=uv

If we plug 1) and 2) into 3) we get:

-b/a=c/a

Multiply both sides by a:

-b=c

Here we have:

a=3

b=-(3k-2)

c=-(k-6)

So we are solving

-b=c for k:

3k-2=-(k-6)

Distribute:

3k-2=-k+6

Add k on both sides:

4k-2=6

Add 2 on both side:

4k=8

Divide both sides by 4:

k=2

Let's check:

3x^2-(3k-2)x-(k-6) \text{ with }k=2:

3x^2-(3\cdot 2-2)x-(2-6)

3x^2-4x+4

I'm going to solve 3x^2-4x+4=0 for x using the quadratic formula:

\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\frac{4\pm \sqrt{(-4)^2-4(3)(4)}}{2(3)}

\frac{4\pm \sqrt{16-16(3)}}{6}

\frac{4\pm \sqrt{16}\sqrt{1-(3)}}{6}

\frac{4\pm 4\sqrt{-2}}{6}

\frac{2\pm 2\sqrt{-2}}{3}

\frac{2\pm 2i\sqrt{2}}{3}

Let's see if uv=u+v holds.

uv=\frac{2+2i\sqrt{2}}{3} \cdot \frac{2-2i\sqrt{2}}{3}

Keep in mind you are multiplying conjugates:

uv=\frac{1}{9}(4-4i^2(2))

uv=\frac{1}{9}(4+4(2))

uv=\frac{12}{9}=\frac{4}{3}

Let's see what u+v is now:

u+v=\frac{2+2i\sqrt{2}}{3}+\frac{2-2i\sqrt{2}}{3}

u+v=\frac{2}{3}+\frac{2}{3}=\frac{4}{3}

We have confirmed uv=u+v for k=2.

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ohaa [14]
<h2>Answer:</h2><h2>-4</h2><h2 /><h2>Hope this helps!!</h2>
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3 years ago
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