All categories

2 years ago

8 times 51 show your work

Mathematics

1 answer:

SCORPION-xisa [38]2 years ago

4 0

8x51. 8x50=? 8x5=40. 40 x 10 = 400 8x50=400. 8x1=8 400+8=408.

You might be interested in

A wise man once said “400 reduced by 3 times my age is 244.” How old is the wise man?

Readme [11.4K]

Answer:

52 years old

Step-by-step explanation:

let this mans age be a

400-3a=244

400=244+3a

400-244=3a

156=3a

a=52

7 0

3 years ago

Read 2 more answers

30 inches to 3 inches

kondor19780726 [428]

Answer:

I believe its 3

Step-by-step explanation:

5 0

2 years ago

Which ratio represents the tangent of an angle?

azamat

Answer:

option d.opposite / adjacent

Step-by-step explanation:

opposite /adjacent ratio represents the tangent of an angle .

hope it is helpful to you ☺️

3 0

2 years ago

Read 2 more answers

Two parallel lines are crossed by a transversal. If measure of angle 6 is 123.5 then measure of angle 1 is.....

GalinKa [24]

The picture in the attached figure

we know that
m ∠ 6=123.5°

m ∠6+m∠2=180°-------> by supplementary angles
so
m∠2=180-m∠6-----> 180-123.5------> m∠2=56.5°

m∠1=m∠2--------> by corresponding angles
so
m∠1=56.5°

the answer is
m∠1 is 56.5°

7 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:

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

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