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3 years ago

Find all solutions to the equation. 2sin theta - squareroot 3 = 0

Mathematics

1 answer:

Xelga [282]3 years ago

8 0

Answer:

$\theta$ =2mπ + π/3 for m ∈ Z.

Step-by-step explanation:

Given the equation $2sin\theta - \sqrt{3} = 0$ , we are to find all the values of $\theta$ that satisfies the equation.

$2sin\theta - \sqrt{3} = 0\\\\2sin\theta = \sqrt{3} \\\\sin\theta = \sqrt{3}/2 \\\\\theta = sin{-1} \sqrt{3}/2 \\\\\theta = 60^0$

General solution for sin $\theta$ is $\theta$ = nπ + (-1)ⁿ ∝, where n ∈ Z.

If n is an even number say 2m, then $\theta$ = (2m)π + ∝ where ∝ = 60° = π/3

Hence, the general solution to the equation will be $\theta$ = 2mπ + π/3 for m ∈ Z.

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5x3 + 14x2 + 9x <br> help

Cerrena [4.2K]

<h3>Answer: x(x+1)(5x+9) </h3>

===================================================

Work Shown:

5x^3 + 14x^2 + 9x

x( 5x^2 + 14x + 9 )

To factor 5x^2 + 14x + 9, we could use the AC method and guess and check our way to getting the correct result.

A better way in my opinion is to solve 5x^2 + 14x + 9 = 0 through the quadratic formula

$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-(14)\pm\sqrt{(14)^2-4(5)(9)}}{2(5)}\\\\x = \frac{-14\pm\sqrt{16}}{10}\\\\x = \frac{-14\pm4}{10}\\\\x = \frac{-14+4}{10} \ \text{ or } \ x = \frac{-14-4}{10}\\\\x = \frac{-10}{10} \ \text{ or } \ x = \frac{-18}{10}\\\\x = -1 \ \text{ or } \ x = \frac{-9}{5}\\\\$

Then use those two solutions to find the factorization

x = -1 or x = -9/5

x+1 = 0 or 5x = -9

x+1 = 0 or 5x+9 = 0

(x+1)(5x+9) = 0

So we have shown that 5x^2 + 14x + 9 factors to (x+1)(5x+9)

-----------

Overall,

5x^3 + 14x^2 + 9x

factors to

x(x+1)(5x+9)

6 0

3 years ago

Polygon ABCD is plotted on a coordinate plane and then rotated 90° clockwise about point C to form polygon A′B′C′D′. Match each

horrorfan [7]

The vertices of ABCD after 90 degrees clockwise rotation about point C are: A' (0,6), B' (3,7), C' (4,6) and D' (4,3)

<h3>How to match the vertices of the polygon?</h3>

The image of the polygon ABCD is not given; however, the question can still be answered because the coordinates are known.

The vertices of polygon ABCD are given as:

A = (4, 6)

B = (5, 3)

C = (4, 2)

D = (1, 2)

The rule of rotation about point C is:

(x,y) = (a + b - y, x + b - a)

Where:

(a, b) = (4, 2) --- the point of rotation.

So, we have:

(x,y) = (4 + 2 - y, x + 4 - 2)

(x,y) = (6 - y, x + 2)

The above means that:

A' = (6 - 6, 4 + 2) = (0,6)

B' = (6 - 3, 5 + 2) = (3,7)

C' = (6 - 2, 4 + 2) = (4,6)

D' = (6 - 2, 1 + 2) = (4,3)

Hence, the image of the rotation and their vertices (i.e. coordinates) are:

A' = (0,6)

B' = (3,7)

C' = (4,6)

D' = (4,3)

Read more about rotation at:

brainly.com/question/4289712

#SPJ1

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Hopefully this helps

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Is the line through points P(2,-9) and Q(6, -13) perpendicular to the line through points R(5,-1) and

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