All categories

3 years ago

What are the period and phase shift for f(x) = −4 tan(x − π)?

Mathematics

1 answer:

777dan777 [17]3 years ago

8 0

This is the answer Period: π; phase shift: x = π

You might be interested in

If a phone cost $224 what would it cost if there was a 7% sales tax.

Annette [7]

Answer:

$239.68

Step-by-step explanation:

Given

Cost = $224
Sales Tax = 7%

Solving

Cost w/ Tax = Cost (1 + Tax)
Cost w/ Tax = 224 (1 + 0.07)
Cost w/ Tax = 224 * 1.07
Cost w/ Tax = $239.68

5 0

2 years ago

Mark spins a penny 100 times and it lands head side up 41 times. Explain why Mark's experimental probability may be different fr

astra-53 [7]

Answer:

Hey there!

This is because the theoretical probability is different from the experimental probability. The theoretical probability is what is expected, but the experimental probability is what actually happens.

Hope this helps :)

4 0

3 years ago

Read 2 more answers

Which solution set is graphed on the number line? 4-3-210

maw [93]

C is your answer because the circle is open and not colored.

6 0

3 years ago

Read 2 more answers

HELP ASAP PLEASEEE.

Troyanec [42]

Please mark Brainliest. Plz.

Answer:

x^2 = 25 x = 5, -5

2x^2 = 98 x = 7, -7

x^2 + 64 = 0 x = 8i, -8i

9x^2 - 16 = 0 x = 4/3, -4/3

x^2 + 9 = 25 x = 4, -4

(x-2)^2 = 25 x = 7, -3

(x-2)^2 +9 = 25 x = 6, -2

4(x-2)^2 +9 = 25 x = 4, 0

8 0

2 years ago

A) Let X be a random variable that can assume only positive integer values, and assume its probability function is P(X -n) A/3^n

kramer

Answer:

a) The value of A = 2

b) The value of $B = \dfrac{1}{12}$

Step-by-step explanation:

Given that:

X should be the random variable that assumes only positive integer values.

The probability function; $P[X = n] = \dfrac{A}{3^n}$ for some constant A and n ≥ 1.

Then, let $\sum \limits ^{\infty}_{n =1} P[X =n] = 1$

This implies that:

$A \sum \limits ^{\infty}_{n =1} \dfrac{1}{3^n}= 1$

$A \times \dfrac{\dfrac{1}{3}}{1 - \dfrac{1}{3}} = 1$

$A \times \dfrac{\dfrac{1}{3}}{\dfrac{2}{3}} = 1$

$A \times \dfrac{1}{2}=1$

A = 2

Thus, the value of A = 2

Suppose X represents a e constant A (n> 1). Find A.

b) Let X be a continuous random variable that can assume values between 0 and 3

Then, the density function of x is:

$f_x(x) = \left \{ {{B(x^2+1)} \ \ \ 0 \le x \le 3 \ \ \ \atop {0} \ \ \ otherwise} \right.$

where; B is constant.

Then, using the property of the probability density function:

$\int ^3_0 \ B (x^2+1 ) \ dx = 1\\$

Taking the integral, we have:

$B \Big [\dfrac{x^3}{3} +x \Big ]^3_0 = 1$

$B \Big [\dfrac{3^3}{3} +3 \Big ]= 1$

$B \Big [\dfrac{27}{3} +3 \Big ] = 1$

B [ 9 +3 ] = 1

B [ 12 ] = 1

Divide both sides by 12

$B = \dfrac{1}{12}$

5 0

3 years ago