Simplify the expression 2.1(3.7 +5.7u) using the Distributive Property.<br>PLEASE HELP! :(

What is the arc length of ACB?

Debora [2.8K]

Answer:

C. 82.43

Step-by-step explanation:

acb's angle is 315 degrees because

360 - 45 = 315

arc length is in radians

arc length is radius times (angle times (pi÷180))

arc length is 15 times (315 times (pi÷180))

arc length is 82.43

cuemathcomgeometry

4 0

2 years ago

Solve dis attachment and show all work ( I got it all wrong and I want to know how to solve it )

DedPeter [7]

(a) First find the intersections of $y=e^{2x-x^2}$ and $y=2$ :

$2=e^{2x-x^2}\implies \ln2=2x-x^2\implies x=1\pm\sqrt{1-\ln2}$

So the area of $R$ is given by

$\displaystyle\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\left(e^{2x-x^2}-2\right)\,\mathrm dx$

If you're not familiar with the error function $\mathrm{erf}(x)$ , then you will not be able to find an exact answer. Fortunately, I see this is a question on a calculator based exam, so you can use whatever built-in function you have on your calculator to evaluate the integral. You should get something around 0.5141.

(b) Find the intersections of the line $y=1$ with $y=e^{2x-x^2}$ .

$1=e^{2x-x^2}\implies 0=2x-x^2\implies x=0,x=2$

So the area of $S$ is given by

$\displaystyle\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}(2-1)\,\mathrm dx+\int_{1+\sqrt{1-\ln2}}^2\left(e^{2x-x^2}-1\right)\,\mathrm dx$
$\displaystyle=2\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\mathrm dx$

which is approximately 1.546.

(c) The easiest method for finding the volume of the solid of revolution is via the disk method. Each cross-section of the solid is a circle with radius perpendicular to the x-axis, determined by the vertical distance from the curve $y=e^{2x-x^2}$ and the line $y=1$ , or $e^{2x-x^2}-1$ . The area of any such circle is $\pi$ times the square of its radius. Since the curve intersects the axis of revolution at $x=0$ and $x=2$ , the volume would be given by

$\displaystyle\pi\int_0^2\left(e^{2x-x^2}-1\right)^2\,\mathrm dx$

5 0

3 years ago

Part C <br> Using the rise and run from part B, what is the slope of the line?

Leni [432]

Hey there, Delilah
What's it like in New York city?
I'm a thousand miles away
But, girl, tonight you look so pretty
Yes, you do
Time square can't shine as bright as you
I swear, it's true
Hey there, Delilah
Don't you worry about the distance
I'm right there if you get lonely
Give this song another listen
Close your eyes
Listen to my voice, it's my disguise
I'm by your side
Oh, it's what you do to me
Oh, it's what you do to me
Oh, it's what you do to me
Oh, it's what you do to me
What you do to me
Hey there, Delilah
I know times are gettin' hard
But just believe me, girl
Someday I'll pay the bills with this guitar
We'll have it good
We'll have the life we knew we would
My word is good
Hey there, Delilah
I've got so much left to say
If every simple song I wrote to you
Would take your breath away
I'd write it all
Even more in love with me you'd fall
We'd have it all
Oh, it's what you do to me
Oh, it's what you do to me
Oh, it's what you do to me
Oh, it's what you do to me
A thousand miles seems pretty far
But they've got planes and trains and cars
I'd walk to you if I had no other way
Our friends would all make fun of us
And we'll just laugh along because we know
That none of them have felt this way
Delilah, I can promise you
That by the time we get through
The world will never ever be the same
And you're to blame
Hey there, Delilah
You be good, and don't you miss me
Two more years and you'll be done with school
And I'll be makin' history like I do
You'll know it's all because of you
We can do whatever we want to
Hey there, Delilah, here's to you
This ones for you
Oh, it's what you do to me
Oh, it's what you do to me
Oh, it's what you do to me
Oh, it's what you do to me
What you do to me
Oh, whoa, whoa
Oh whoa, oh
Oh, oh

4 0

3 years ago

What is the equation to find the slope <br> m=

DiKsa [7]

Answer:

$m=\frac{y2-y1}{x2-x1}$