What is the best estimate for the answer to 321.786 + 240.273?

HELP ME PLEASE !!!!!?!!!!!?!!!!!!!!!!!

motikmotik

Answer:

First Graph (The left-most graph)

Step-by-step explanation:

5 0

3 years ago

If the price of a bag of Ritz peanut butter crackers increases from $0.70 to $0.80, what it the percent increase?

Shtirlitz [24]

Answer:

14%

Step-by-step explanation:

To get the percent increase divide 0.8 by 0.7

0.8/0.7 ≈ 1.14285714286

So the percent increase would by about 14 percent

3 0

2 years ago

A trapezoid is a quadrilateral with exactly one pair of parallel sides A) true B)False

Alenkinab [10]

The answer is true to your question

7 0

3 years ago

Read 2 more answers

How do you measure the perimeter of a shape

Advocard [28]

Perimeter is a measure of the area or distance around a two-dimensional shape. On a rectangle, for instance, the perimeter is the total length of the rectangle's outline, including the two widthwise borders and the two lengthwise borders.

To determine the perimeter of a shape, therefore, you add together all the dimensions that make up the shape's outer edge. Being able to find the perimeter of a shape has many applications in the real world. Say, for example, that you wanted to build a fence around your backyard. In order to purchase materials, you need to know how much fencing you'll need, and to determine that you have to figure out the perimeter of the area you want to fence in.

8 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