Consider the line y = -

1. Find the phase shift of the function y = 5cos(2x + pi/2).

kolezko [41]

Answer:

see explanation

Step-by-step explanation:

1

The cosine function in standard form is

y = acos(bx + c)

where a is the amplitude, period = $\frac{2\pi }{b}$ and

phase shift = - $\frac{c}{b}$

here b = 2 and c = $\frac{\pi }{2}$ , thus

phase shift = - $\frac{\frac{\pi }{2} }{2}$ = - $\frac{\pi }{4}$

2

the amplitude = | a |

which has a maximum of a and a minimum of - a

y = 4cosx ← has a maximum value of 4

5 0

3 years ago

Luke is 5 years younger than 3 times in Sydney's age in the situation what does 3s represnt

Fudgin [204]

3s is 3 times sydneys age. so the equation would be 3s-5

8 0

3 years ago

Physicians at a clinic gave what they thought were drugs to 860 patients. Although the doctors later learned that the drugs were

Alexus [3.1K]

Answer:

Z=2.36

P=0.0091

Step-by-step explanation:

The hypothesis is:

π > 0.55 which is the population proportion

The formula to get the test statistic is

Zstat = P-π/√π(1-π)/n

P =507.4/860 = 0.59 (sample proportion )

n =860 (sample size)

π =0.55(population porportion)

Zstat = 0.59-0.55/√0.55(1-0.55)/860

Zstat =2.36

From Z table the probability of Z score of 2.36 is 0.9909

To calculate for P- value will then be

P= 1 - 0.9909

P= 0.0091

4 0

3 years ago

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Jlenok [28]

Answer:

$x>2$

Step-by-step explanation:

When given the following inequality;

$(x^2+x-3):(x^2-4)\geq1$

Rewrite in a fractional form so that it is easier to work with. Remember, a ratio is another way of expressing a fraction where the first term is the numerator (value over the fraction) and the second is the denominator(value under the fraction);

$\frac{x^2+x-3}{x^2-4}\geq1$

Now bring all of the terms to one side so that the other side is just a zero, use the idea of inverse operations to achieve this:

$\frac{x^2+x-3}{x^2-4}-1\geq0$

Convert the (1) to have the like denominator as the other term on the left side. Keep in mind, any term over itself is equal to (1);

$\frac{x^2+x-3}{x^2-4}-\frac{x^2-4}{x^2-4}\geq0$

Perform the operation on the other side distribute the negative sign and combine like terms;

$\frac{(x^2+x-3)-(x^2-4)}{x^2-4}\geq0\\\\\frac{x^2+x-3-x^2+4}{x^2-4}\geq0\\\\\frac{x+1}{x^2-4}\geq0$

Factor the equation so that one can find the intervales where the inequality is true;

$\frac{x+1}{(x-2)(x+2)}\geq0$

Solve to find the intervales when the equation is true. These intervales are the spaces between the zeros. The zeros of the inequality can be found using the zero product property (which states that any number times zero equals zero), these zeros are as follows;

$-1, 2, -2$

Therefore the intervales are the following, remember, the denominator cannot be zero, therefore some zeros are not included in the domain

$x\leq-2\\-2$

Substitute a value in these intervales to find out if the inequality is positive or negative, if it is positive then the interval is a solution, if it is negative then it is not a solution. This is because the inequality is greater than or equal to zero;

$x\leq-2$ -> negative