What is the overall charge of phosphide

If f(x) = |x| and g(x) = |x| − 4, which transformation is applied to f(x) to get g(x)

velikii [3]

The graphs have precisely the same shape, but that of g(x) is that of f(x) translated 4 units DOWN.

7 0

3 years ago

Please answer my question .q bo 4 d and e

sdas [7]

4 (e)
sin^8 x - cos^8 x
= (sin^4 x + cos^4 x)(sin^4 x - cos^4 x)
= (sin^4 x + cos^4 x)(sin^2 x - cos^2x)(sin^2 x +cos^2 x)
= (sin^4x + cos^4 x)( sin^2 x- cos^2 x)

Sorry I cant do 4 (d).

5 0

3 years ago

The graph below shows the velocity f(t) of a runner during a certain time interval:

Snezhnost [94]

When graphs are plotted, graphs as in the .jpeg image in attachment are obtained.
First graph is line going through ordered pair (point) (0,6) and pair (6,8).
The second graph is line going through ordered pair (6,8) and pair (8,0).
Intercept of these graphs is point (ordered pair) (6,8).
You see from the .jpeg image that the following is true:
Initial velocity of the runner is 6 meters per second (runner starts with this velocity), for a while he runs, velocity grows, then in the 6th second (time=6), runner starts to slow down and velocity starts to decrease, and runner stops totally at time = 8.
So runner stops after 8 seconds.
The 4th claim (sentence) is correct.

3 0

3 years ago

Read 2 more answers

Which relation is a function?

Fudgin [204]

It's the one second below the question because it does not have two outcomes for x. For example in a relation x could result in y = 2 or y = -2.

Hope this helps! :)

6 0

2 years ago

Read 2 more answers

Terri Vogel, an amateur motorcycle racer, averages 129.71 seconds per 2.5 mile lap (in a 7 lap race) with a standard deviation o

Vikki [24]

Answer:

(a) The percent of her laps that are completed in less than 130 seconds is 55%.

(b) The fastest 3% of her laps are under 125.42 seconds.

(c) The middle 80% of her laps are from 126.80 seconds to 132.63 seconds.

Step-by-step explanation:

The random variable X is defined as the number of seconds for a randomly selected lap.

The random variable X is normally distributed with mean, μ = 129.71 seconds and standard deviation, σ = 2.28 seconds.

Thus, $X\sim N(129.71,\ 2.28^{2})$ .

(a)

Compute the probability that a lap is completes in less than 130 seconds as follows:

$P(X$

$=P(Z$

The percentage is, 0.55 × 100 = 55%.

Thus, the percent of her laps that are completed in less than 130 seconds is 55%.

(b)

Let x represents the 3rd percentile.

That is, P (X < x) = 0.03.

⇒ P (Z < z) = 0.03

The value of z for the above probability is:

z = -1.88

Compute the value of x as follows:

$z=\frac{x-\mu}{\sigma}\\-1.88=\frac{x-129.71}{2.28}\\x=129.71-(1.88\times 2.28)\\x=125.4236\\x\approx 125.42$

Thus, the fastest 3% of her laps are under 125.42 seconds.

(c)

Let x₁ and x₂ be the values between which the middle 80% of the distribution lie.

That is,

$P(x_{1}$

The value of z for the above probability is:

z = 1.28

Compute the values of x₁ and x₂ as follows:

$-z=\frac{x_{1}-\mu}{\sigma}\\-1.28=\frac{x_{1}-129.71}{2.28}\\x_{1}=129.71-(1.28\times 2.28)\\x=126.7916\\x\approx 126.80$ $z=\frac{x_{2}-\mu}{\sigma}\\1.28=\frac{x_{2}-129.71}{2.28}\\x_{2}=129.71+(1.28\times 2.28)\\x=132.6284\\x\approx 132.63$

Thus, the middle 80% of her laps are from 126.80 seconds to 132.63 seconds.

5 0

3 years ago