What are the domain range and asymptote of h(x)=(1.4)^x+5

Set G consists of: G={4, 8, 12, 16, 20, 24, 28, 32, 36,...} Set F represents the perfect squares: F={1, 4, 9, 16, 25, 36, 49, 64, 81, 100...} Within set F, the numbers 4, 16, 36, 64, and 100 are multiples of 4. The result is: {4, 16, 36, 64, 100}.

To tackle this sinusoidal question, we begin with the following: Using the formula; g(t)=offset+A*sin[(2πt)/T+Delay] According to sinusoidal theory, the duration from trough to crest is typically half of the wave's period. Here, T=2.5 The peak magnitude is calculated as: Trough-Crest=2.1-1.5=0.6 m amplitude=1/2(Trough-Crest)=1/2*0.6=0.3 The offset from the center of the circle becomes 0.3+1.5=1.8 As the delay is at -π/2, the wave will commence at the trough at [time,t=0]. Plugging these values into the formula gives: g(t)=1.8+(0.3)sin[(2*π*t)/2.5]-π/2] g(t)=1.8+0.3sin[(0.8πt)/T-π/2]

Answer:

Approximately 59 stacked cups.

Step-by-step explanation:

We have the following measurements:

Height of one cup = 12.5 cm,

Height of two cups stacked = 14 cm,

Height of three cups stacked = 15.5 cm,

...and so on.

This situation can be described by an arithmetic sequence,

12.5, 14, 15.5,....

The first term is defined as a = 12.5,

with a common difference of d = 1.5 cm.

Thus, the height of x stacked cups is given by

As per the problem,

h(x) = 200

⇒ 1.5x + 11 = 200

⇒ 1.5x = 189

⇒ x = 59.3333333333 ≈ 59.

Therefore, you will need approximately 59 stacked cups.

Answer:d

Step-by-step explanation:

Answer:

Step-by-step explanation:

Player A has a red marble and a blue marble, while Player B also has a red marble and a blue marble.

Therefore, the probability of selecting one marble is equal, at 0.5.

Due to the independence of A and B's choices, the joint event is calculated by multiplying the probabilities.

Let A represent the amount that player A wins.

If both players select one marble, the sample space can be considered as

             (R,R)  (R,B)  (B,R) (B,B)

Probability     0.25  0.25  0.25  0.25

A's winnings          3       -2     -2       1

E(A)      0.75   -0.5    -0.5   0.25   =    0

Thus, the game is even, offering equal expected values for both A and B.

It does not influence the outcome whether you are A or B.