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2 years ago

Which function has a domain of all real numbers

Mathematics

2 answers:

Hoochie [10]2 years ago

8 0

Answer: where are the numbers

Step-by-step explanation:

Pachacha [2.7K]2 years ago

4 0

Answer:

Most of the functions we have studied in Algebra I are defined for all real numbers. This domain is denoted . For example, the domain of f (x) = 2x + 5 is , because f (x) is defined for all real numbers x; that is, we can find f (x) for all real numbers x.

Step-by-step explanation:

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Solve r = 4/3(p-q) for p

ddd [48]

Answer:

q+3/4r=p

Step-by-step explanation:

r=4/3(p-q)

Distribute the 4/3

r=4/3p-4/3q

Add 4/3q to each side

4/3q+r=4/3p

Multiply ALL variables by 3/4 (undoes the 4/3)

q+3/4r=p

6 0

2 years ago

If LM=22 and MN=15,find LN

ziro4ka [17]

Answer: 37 you just add the two in that case

8 0

2 years ago

Les baked 48 cookies for the bake sale at school, and sold 36 of them. in fraction form, how many cookies did Les have left?

inna [77]

Answer:

Les sold 36 of 48.

36/48

Step-by-step explanation:

36 over 48 reduces to 3/4.

she sold 3/4 of the cookies.

4 0

2 years ago

Read 2 more answers

3. You can ride your bike 1/5 of a mile per minute. If it takes you 3 1/3 minutes to get

Allushta [10]

Answer:

2/3

.......................

6 0

2 years ago

Set up the integral that represents the arc length of the curve f(x) = ln(x) + 5 on [1, 3], and then use Simpson's Rule with n =

marta [7]

Answer:

The integral for the arc of length is:

$\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx$

By using Simpon’s rule we get: 1.5355453

And using technology we get: 2.3020

The approximation is about 33% smaller than the exact result.

Explanation:

The formula for the length of arc of the function f(x) in the interval [a,b] is:

$\displaystyle\int_a^b \sqrt{1+[f'(x)]^2}dx$

We need the derivative of the function:

$f'(x)=\frac{1}{x}$

And we need it squared:

$[f'(x)]^2=\frac{1}{x^2}$

Then the integral is:

$\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx$

Now, the Simposn’s rule with n=4 is:

$\displaystyle\int_a^b g(x)}dx\approx\frac{\Delta x}{3}\left( g(a)+4g(a+\Delta x)+2g(a+2\Delta x) +4g(a+3\Delta x)+g(b) \right)$

In this problem:

$a=1,b=3,n=4, \displaystyle\Delta x=\frac{b-a}{n}=\frac{2}{4}=\frac{1}{2},g(x)= \sqrt{1+\frac{1}{x^2}}$

So, the Simposn’s rule formula becomes:

$\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx\\\approx \frac{\frac{1}{3}}{3}\left( \sqrt{1+\frac{1}{1^2}} +4\sqrt{1+\frac{1}{\left(1+\frac{1}{2}\right)^2}} +2\sqrt{1+\frac{1}{\left(1+\frac{2}{2}\right)^2}} +4\sqrt{1+\frac{1}{\left(1+\frac{3}{2}\right)^2}} +\sqrt{1+\frac{1}{3^2}} \right)$

Then simplifying a bit:

$\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx \approx \frac{1}{9}\left( \sqrt{1+\frac{1}{1^2}} +4\sqrt{1+\frac{1}{\left(\frac{3}{2}\right)^2}} +2\sqrt{1+\frac{1}{\left(2\right)^2}} +4\sqrt{1+\frac{1}{\left(\frac{5}{2}\right)^2}} +\sqrt{1+\frac{1}{3^2}} \right)$

Then we just do those computations and we finally get the approximation via Simposn's rule:

$\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx\approx 1.5355453$

While when we do the integral by using technology we get: 2.3020.

The approximation with Simpon’s rule is close but about 33% smaller:

$\displaystyle\frac{2.3020-1.5355453}{2.3020}\cdot100\%\approx 33\%$

8 0

2 years ago