Rewrite 14 1/3 as an improper fraction

There are 5 less oranges on the tree today than there was yesterday Plz help

topjm [15]

That’s all ? is there like no other explanation to make it understood better ?

8 0

3 years ago

Read 2 more answers

Select all of the symbols that would make the comparison true. -7__-4

Marysya12 [62]

Answer:uggvjyugjyrftkufiituu

Step-by-step explanation:

gggggggggggggggggggggggggggggggggggggggggggggggggg

3 0

3 years ago

Please help___________

kramer

Answer:

Step-by-step explanation:

Let's do the obvious things first.

Right rectangle

w = 11

L = 17

Formula

Area = L * W

Area = 11 * 17

Area = 187

Left rectangle

W = 8

L = 11

Area = 8 * 11= 88

Middle Rectangle

W = 11

L = 15

Area = 11 * 15 165

Now we come to the triangles. It's not obvious what to do with them. You have to infer that one of the lets is 8 and the other is 15. You get the 8 from the width of the left rectangle (see above.)

There are 2 triangles

Area = 2 * 1/2 * l1 * l2

Leg 1 = 8

Leg2 = 15

Area = 2 * 1/2 (8 * 15)

Area = 120

Total = 120 + 165 + 88 + 187 = 560

Unless you have an E answer that is 560, my answer does not agree with any of the given answers. I would suggest that you ask your teacher how it is done. My method is correct.

4 0

3 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:

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