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3 years ago

10

Is equivalent to 4sqrt(27) + 3sqrt(12) 18sqrt(3) 7sqrt(3) 5sqrt(3) 7sqrt(39)

1 answer:

Angelina_Jolie [31]3 years ago

3 0

When you say sweet what does it equal

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Describe and correct the error in finding the measure of the angle

DerKrebs [107]

Sorry, it's late, and I'm a bad explainer.

The error is adding (2x-12) with x and 30. This is wrong because you are adding the angles inside the triangle and you are assuming that (2x - 12) is the unlabeled angle INSIDE the triangle, when it is the exterior angle/outside of the triangle.

A straight line is also 180°.

(2x - 12) + ? = 180

30 + x + ? = 180

If you look at the equations, and put parentheses around 30 + x, (30 + x) and (2x - 12) should be the SAME NUMBER. So you could set them equal to each other to find x. (or you could also look at the picture and see that they both need/are missing the same angle)

2x - 12 = 30 + x

x = 42

Now you plug 42 into the exterior angle equation

2(42) - 12 = 84 - 12 = 72°

5 0

3 years ago

This is due tomorrow if someone could help me that would be great <3

kifflom [539]

<u>Hope this helped!</u>

<u>First Row</u>

$\frac{5}{18}$

$\frac{10}{21}$

$\frac{9}{40}$

6

5 $\frac{1}{3}$

(this one is already done)

<u>Second Row</u>

4 $\frac{4}{9}$

5 $\frac{1}{16}$

$\frac{13}{18}$

6 $\frac{1}{4}$

11 $\frac{2}{3}$

5 0

3 years ago

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.

4 0

3 years ago

find the number of possible combinations choose one of three books and one of eight CDs to bring on a bus trip

SVEN [57.7K]

There is a possibility of 24 combinations I believe

6 0

4 years ago

Read 2 more answers

What are the zeros of this function?

Hunter-Best [27]

If you need the y zero, you would take the 2 points on the graph of (3, 0) and (6, 0) so the 2 zero’s are 3, and 6.

3 0

2 years ago