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2 years ago

PLEASE HELP ME IM SORRY PLEASE

Mathematics

1 answer:

Gnom [1K]2 years ago

3 0

Hello! (:

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5 less than 3.1 times a number n:

3.1n-5

Four more than the quotient of 12 and a number n:

n/12+4

Hope this helps you!

~SparklingFlower

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A toll-free sales line sold 85 products for every 125 calls in one day. What is the daily success rate of the sales line? A. 0.1

IgorLugansk [536]

Answer:

C. 0.68

Step-by-step explanation:

Given;

number of products sold in a day by toll-free sales line = 85 products

number of calls in a day = 125

The daily success rate of the sales line is given by the ratio of the total products in a day to total number of calls in a day.

The daily success rate of the sales line = total products sold / number of calls

The daily success rate of the sales line = 85 / 125

The daily success rate of the sales line = 0.68

Therefore, the daily success rate of the sales line is 0.68

7 0

2 years ago

Can someone pls help im new to this

Serggg [28]

Answer:

Step-by-step explanation:

$\frac{-10}{-2} \\\\5$

3 0

3 years ago

please help me solve this. ill award brainliest! please just explain how to solve and what it should be.

Galina-37 [17]

Answer:

3 - that is if all of those say 7 because some have a line through and some dont.

Step-by-step explanation:

Download photomath. You take a photo of the math problem and it gives the answer plus steps to solve.

4 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

3 years ago

Please help! it’s urgent

ycow [4]

Step-by-step explanation:

${9}^{ \frac{3}{2} } \times {27}^{ \frac{1}{2} } = {(3)}^{ \frac{3}{2} \times 2 } \times {(3)}^{ \frac{1}{2} \times 3} \\ = {3}^{3} \times {3}^{ \frac{3}{2} } \\ = {3}^{3 + \frac{3}{2} } \\ = {3}^{4\frac{1}{2} } \\ = {3}^{ \frac{9}{2} }$

8 0

2 years ago