All categories

3 years ago

4 sin ^ 2 x- 4 sin x + 1 = 0

1 answer:

8 0

$4\sin^2x-4\sin x+1=(2\sin x-1)^2=0\implies \sin x=\dfrac12$

This happens for $x=\dfrac\pi6+2n\pi$ and $x=\dfrac{5\pi}6+2n\pi$ where $n$ is any integer.

You might be interested in

A. If a bus travels 6km in to minutes, what distance does it travel in Isecondo

mina [271]

Answer:

60 second

Step-by-step explanation:

6km into minutes

1 minutes is 60 second

7 0

3 years ago

Read 2 more answers

Two planes are flying in opposite directions, away from each other, one with the speed of 800 km per hour and the other with the

allsm [11]

1640 km every hour is the answer

3 0

2 years ago

Read 2 more answers

I can’t figure out how to use the zeros in the polynomial. Please explain

Temka [501]

Answer:

a) $P (x) = (x + 3) (x-1) (x-4)$

b) $P (x) = (2x + 5) (5x - 4) (x-6)$

c) $P (x) = (x-3) (x-1) (x-4) (x + 1) ^ 2$

Step-by-step explanation:

For the question a * you need to find a polynomial of degree 3 with zeros in -3, 1 and 4.

This means that the polynomial P(x) must be zero when x = -3, x = 1 and x = 4.

Then write the polynomial in factored form.

$P (x) = (x + 3) (x-1) (x-4)$

Note that this polynomial has degree 3 and is zero at x = -3, x = 1 and x = 4.

For question b, do the same procedure.

Degree: 3

Zeros: -5/2, 4/5, 6.

The factors are

$x = -\frac{5}{2}\\\\x +\frac{5}{2} = 0\\\\(2x +5) = 0$

---------------------------------------

$x =\frac{4}{5}\\\\x-\frac{4}{5} = 0\\\\(5x-4) = 0$

--------------------------------------

$x = 6\\\\(x-6) = 0$

--------------------------------------

$P (x) = (2x + 5) (5x - 4) (x-6)$

Finally for the question c we have

Degree: 5

Zeros: -3, 1, 4, -1

Multiplicity 2 in -1

$x = -3\\\\(x-3) = 0$

--------------------------------------

$x = 1\\\\(x-1) = 0$

--------------------------------------

$x = 4\\\\(x-4) = 0$

----------------------------------------

$x = -1\\\\(x + 1) = 0$

-----------------------------------------

$P (x) = (x-3) (x-1) (x-4) (x + 1) ^ 2$

8 0

2 years ago

Which is a possible turning point for the continuous function f(x)?

babymother [125]

Answer:

We are given coordinates of a continuous function f(x)

(–2, 0)

(0, –2)

(2, –1)

(4, 0).

We need to find the possible turning point for the continuous function.

Note: Turning point is a point on the graph where slope of the curve changes from negative to positive or positive to negative.

A turning point is always lowest or highest point of the curve (where bump of the graph seen).

For the given coordinates we can see that (–2, 0) and (4, 0) coordinates are in a same line, that is on the x-axis.

But the coordinate (0, –2) is the lowest point on the graph.

Therefore, (0, –2) is the turning point for the continuous function given.

hoped this was helpful!

5 0

2 years ago

What are the roots of the equation?

Mariana [72]

Answer:

3x2

Step-by-step explanation:

because it is

4 0

3 years ago