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2 years ago

HELP PLEASE HELPPPPP

Mathematics

1 answer:

ad-work [718]2 years ago

3 0

First find the area of the bigger circle (38^2 x pi since the diameter is 76) and the area of the smaller circle is (19^2 x pi)
So.. 4534.16 (use 3.14 for) - 1133.54 = 3400.62 square yds

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You and your four friends are going to a concert. The total price for the tickets was $118.75 which includes a $10 service fee.

Alex787 [66]

Answer:475

Step-by-step explanation:

118.75 is the cost whitch includes a 10$ service fee

118.75x10=475

8 0

3 years ago

PLZ Answer ASAP!!!

Sliva [168]

Answer:

D. x = 3

Step-by-step explanation:

The only way we could get a line of symmetry through this figure is to use a vertical line. The only way to get symmetry would be to put the vertical line though the point x = 3. When we do that, we can see that the end points of the base are both one unit away from the line of symmetry. The end points I'm talking about are the points at the coordinates (2, 4), and (4, 4)

6 0

3 years ago

I need help. First answer that explains well gets to be marked brainlist. Extra points

Blababa [14]

The empirical probability of a 4 or 5 is
(count of outcome = 4 or 5)/(count of all outcomes) = (44 +40)/240 = 7/20

The problem statement asks you to assume the outcomes will remain in this proportion going forward. That is, the number of outcomes (n) of 4 or 5 in the next 60 rolls of the number cube is expected to satisfy
7/20 = n/60

Multiplying that by 60 gives
n = 60*(7/20) = 21

Selection D is appropriate.

3 0

3 years ago

Does anyone know the simplest radical form for this question

Mkey [24]

Answer:

Step-by-step explanation:

Simplify the radical by breaking up the radicand up into a product of known factors, assuming positive real numbers.

8 0

2 years ago

Read 2 more answers

Use integration by parts to find the integrals in Exercise.<br> x^3 ln x dx.

34kurt

Answer:

$\frac{\text{ln}(x)x^4}{4}-\frac{x^4}{16}+C$ .

Step-by-step explanation:

We have been given an indefinite integral $\int \:x^3\:ln\:x\:dx$ . We are asked to find the value of the integral using integration by parts.

$\int\: u\text{dv}=uv-\int\: v\text{du}$

Let $u=\text{ln}(x)$ , $v'=x^3$ .

Now, we will find du and v as shown below:

$\frac{du}{dx}=\frac{d}{dx}(\text{ln}(x))$

$\frac{du}{dx}=\frac{1}{x}$

$du=\frac{1}{x}dx$

$v=\frac{x^{3+1}}{3+1}=\frac{x^{4}}{4}$

Upon substituting our values in integration by parts formula, we will get:

$\int \:x^3\:\text{ln}\:(x)\:dx=\text{ln}(x)*\frac{x^4}{4}-\int\: \frac{x^4}{4}*\frac{1}{x}dx$

$\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\int\: \frac{x^3}{4}dx$

$\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\frac{1}{4}\int\: x^3dx$

$\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\frac{1}{4}*\frac{x^{3+1}}{3+1}+C$

$\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\frac{1}{4}*\frac{x^4}{4}+C$

$\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\frac{x^4}{16}+C$

Therefore, our required integral would be $\frac{\text{ln}(x)x^4}{4}-\frac{x^4}{16}+C$ .

5 0

2 years ago