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Liono4ka [1.6K]
3 years ago
8

Please help me, it would rlly be appreciated <3​

Mathematics
1 answer:
love history [14]3 years ago
5 0

Answer:

C.

General Formulas and Concepts:

<u>Algebra I</u>

  • Domain is the set of x-values that can be inputted into function f(x)
  • Range is the set of y-values that are outputted by function f(x)

Step-by-step explanation:

<u>Domain</u>

We see from the graph that our x-values span from -3 to 2. Since both are closed dots, they are inclusive:

[-3, 2] or -3 ≤ x ≤ 2

<u>Range</u>

We see from the graph that our y-values span from -2 to 4. Since both are closed dots, they are inclusive:

[-2, 4] or -2 ≤ y ≤ 4

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Choose the true statement about the graph of x &gt; -3.
liberstina [14]

Answer:

b

Step-by-step explanation:

open bc it doesn't include -3

&

to the right bc its greater than -3

6 0
3 years ago
Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis. Verify y
Drupady [299]

Answer:

V = \frac{\pi^2}{8}

V = 1.23245

Step-by-step explanation:

Given

y = \cos 2x

y = 0; x = 0; x = \frac{\pi}{4}

Required

Determine the volume of the solid generated

Using the disk method approach, we have:

V = \pi \int\limits^a_b {R(x)^2} \, dx

Where

y = R(x) = \cos 2x

a = \frac{\pi}{4}; b =0

So:

V = \pi \int\limits^a_b {R(x)^2} \, dx

Where

y = R(x) = \cos 2x

a = \frac{\pi}{4}; b =0

So:

V = \pi \int\limits^a_b {R(x)^2} \, dx

V = \pi \int\limits^{\frac{\pi}{4}}_0 {(\cos 2x)^2} \, dx

V = \pi \int\limits^{\frac{\pi}{4}}_0 {\cos^2 (2x)} \, dx

Apply the following half angle trigonometry identity;

\cos^2(x) = \frac{1}{2}[1 + \cos(2x)]

So, we have:

\cos^2(2x) = \frac{1}{2}[1 + \cos(2*2x)]

\cos^2(2x) = \frac{1}{2}[1 + \cos(4x)]

Open bracket

\cos^2(2x) = \frac{1}{2} + \frac{1}{2}\cos(4x)

So, we have:

V = \pi \int\limits^{\frac{\pi}{4}}_0 {\cos^2 (2x)} \, dx

V = \pi \int\limits^{\frac{\pi}{4}}_0 {[\frac{1}{2} + \frac{1}{2}\cos(4x)]} \, dx

Integrate

V = \pi [\frac{x}{2} + \frac{1}{8}\sin(4x)]\limits^{\frac{\pi}{4}}_0

Expand

V = \pi ([\frac{\frac{\pi}{4}}{2} + \frac{1}{8}\sin(4*\frac{\pi}{4})] - [\frac{0}{2} + \frac{1}{8}\sin(4*0)])

V = \pi ([\frac{\frac{\pi}{4}}{2} + \frac{1}{8}\sin(4*\frac{\pi}{4})] - [0 + 0])

V = \pi ([\frac{\frac{\pi}{4}}{2} + \frac{1}{8}\sin(4*\frac{\pi}{4})])

V = \pi ([{\frac{\pi}{8} + \frac{1}{8}\sin(\pi)])

\sin \pi = 0

So:

V = \pi ([{\frac{\pi}{8} + \frac{1}{8}*0])

V = \pi *[{\frac{\pi}{8}]

V = \frac{\pi^2}{8}

or

V = \frac{3.14^2}{8}

V = 1.23245

4 0
3 years ago
Angie was told to read a novel for English class. She has already read the first 12 pages and can read 10 pages per hour.
Nataly [62]
The equation is y=10x+12
8 0
3 years ago
How do you distribute 9(w+5)=9
Rufina [12.5K]

Answer:

the answer is w = -4

Step-by-step explanation:

6 0
4 years ago
Read 2 more answers
On a multiple-choice test. Abby randomly guesses on all seven questions. Each question
Murrr4er [49]

Answer:

0.173 probability that she gets exactly three questions correct.

Step-by-step explanation:

For each question, there are only two possible outcomes. Either she guesses the correct answer, or she does not. The probability of guessing the correct answer for a question is independent of other questions. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

Seven questions:

This means that n = 7

Each question has four choices.

Abby guesses, which means that p = \frac{1}[4} = 0.25

Find the probability to the nearest thousandth, that Abby gets exactly three questions correct.

This is P(X = 3).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 3) = C_{7,3}.(0.25)^{3}.(0.75)^{4} = 0.173

8 0
3 years ago
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