A coin lands on Heads 200 times. The relative frequency of Heads is 0.4 How many times was the coin thrown?

Answer:

That would depend on his remaining balance prior to purchasing the gaming system.

Answer: The option that is correct is

(C) T0,2(x, y)∘ R0, 270°.

Step-by-step explanation: We need to choose the transformation rule that defines how rectangle PQRS is transformed into P''Q''R''S''.

Observing the graph, we can identify that

the points of rectangle PQRS are P(-3, -5), Q(-2, -5), R(-2, -1) and S(-3, -1).

Conversely, the vertices of rectangle P''Q''R''S'' are P''(-5, 5), Q''(-5, 4), R''(-1, 4) and S(-1, 5).

The rectangle PQRS resides in Quadrant III while P''Q''R''S'' is located in Quadrant II,indicating that the rotation could either be 90° clockwise or 270° counterclockwise around the origin.

Among the provided choices, there is no 90° rotation available, hence we will consider the 270° counterclockwise rotation around the origin.

After executing this rotation, the points transform based on the rule

(x, y) ⇒ (y, -x).

Thus, the vertices of the resulting rectangle P'Q'R'S', after a 270° counterclockwise rotation around the origin, become

P(-3, -5) ⇒ P'(-5, 3),

Q(-2, -5) ⇒ Q'(-5, 2),

R(-2, -1) ⇒ R'(-1, 2),

S(-3, -1) ⇒ S'(-1, 3).

To align the vertices of P'Q'R'S' with those of P''Q''R''S'', we must add 2 units to the y-coordinate of all vertices , leading to

P'(-5, 3) ⇒ P''(-5, 3+2) = P''(-5, 5),

Q'(-5, 2) ⇒ Q''(-5, 2+2) = Q''(-5, 4),

R'(-1, 2) ⇒ R''(-1, 2+2) = R''(-1, 4),

S'(-1, 3) ⇒ S''(-1, 3+2) = S''(-1, 5).

Thus, the required transformation rule becomes

a rotation through 270° counterclockwise around the origin coupled with a translation of (x, y) ⇒ (x, y+2).

Since rigid transformations are documented from right to left, option (C) is indeed correct.

you can set this up with the equation;

x + (x + 42) = 138

start by combining like terms;

2x = 138 - 42

2x = 96

x = 96/2

x = 48

we've found x now plug it back into the original equation.

48 + (48 + 42) = 138

48 + 90 = 138

hope it helped...if you have any concerns just let me know:) 

Answer:

Step-by-step explanation:

The trigonometric expression given is:

11 cos² x +3 sin² x + 6sin x cos x + 5

Alternatively, we can present it as:

(9 cos² x + 2 cos² x) + (2 sin² x + sin² x) + 6sin x cos x + 5

After rearranging, the expression can be rewritten as:

(9 cos² x + sin² x + 6sin x cos x) + (2 cos² x + 2 sin² x) + 5

If we factor the highlighted part similarly to a polynomial:

(9 cos² x + sin² x + 6sin x cos x) + (2 cos² x + 2 sin² x) + 5

This results in:

(3 cos x + sin x)² + 2 (cos² x + sin² x) + 5

Notably, the term in the second set of brackets (cos² x + sin² x) is a well-known trigonometric identity and equals one.

Consequently, the expression simplifies to:

(3 cos x + sin x)² + 7

The maximum and minimum values of the entire expression rely on the max and min values of (3 cos x + sin x), which follows the format (a cos x + b sin x).

The max and min values can be easily determined.

I've included a screenshot from related material below:

Here, a=3 and b=1, thus, R= √10

As the cosine value for any angle ranges between -1 and 1, the value of cos(x − α) will also fall within this range.

This means the max and min for (a cos x + b sin x) will be -R to R, and all resultant values will be between those limits.

In this case, we observe that it falls between (-√10) and √10.

Returning to our original expression:

(3 cos x + sin x)² + 7

The bracketed term ranges between (-√10) and √10.

Nevertheless, since squaring a negative value yields a positive result, we cannot use a negative value for determining the minimum. The minimum occurs at the lowest non-negative value, which is zero.

Thus, the minimum value is:

(0)² + 7 = 7

And the maximum is:

(√10)² + 7 = 17