All categories

3 years ago

33 47 51 53 59 59 83 99 what is the middle

Mathematics

2 answers:

Andre45 [30]3 years ago

4 0

59 is the answer, your welcome gfjgvffb

Jobisdone [24]3 years ago

3 0

Answer:

Step-by-step explanation:

You might be interested in

The Sears Tower in Chicago is 1450 feet high. A model of the tower is 48 inches tall. What is the ratio of the

Leni [432]

Answer: choice a.

Step-by-step explanation:

1 feet = 12 inches.

Sears tower = 1450 * 12 = 17400 inches.

Now,

48 : 17400

= 2 : 725

5 0

3 years ago

Jeremy will roll a number cube, numbered 1-6, twice. What is the probability of rolling an even number than the number 3?

andreyandreev [35.5K]

There are 6 possible outcomes on a dice. Meaning when you roll it it could be any of them. Because 3 is one of the 6 the answer would be 1/6

3 0

3 years ago

I need 20 points reallly quick

tekilochka [14]

Answer my question there is 25 points

5 0

3 years ago

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

3 years ago

$12,000 at 13% for 2years

Mademuasel [1]

Answer:

The interest is $3120 and the amount is $15120.

Step-by-step explanation:

Hope this helped

5 0

3 years ago