Ask question

Ask question

All categories

Law Incorporation [45]

3 years ago

12

I will be honest I'm having a major brain fart.. What's the line called and what's it do again? it's separating numbers.. Looks

like this
29
---
3

I know I'm stupid for this one... I've been up for 30 hours studying and now I can't remember this.. I need sleep but can't till i figure this out now...

1 answer:

Alla [95]3 years ago

5 0

Answer:

fraction bar. separates numerator and denominator in a fraction

You might be interested in

Need help I don’t know how to do it

liberstina [14]

First attempt to break the top portion up so the negative signs don’t confuse you.

(2 x -2) x (-2 x -4)/ 2 x 2

= (-4) x (8)/ 4

= -32/4

= -8

*Remember that multiplying two negative numbers give you a positive number

8 0

2 years ago

Find the equation of a line passing through the points (2,6) and (-2,-10)

navik [9.2K]

$\bf (\stackrel{x_1}{2}~,~\stackrel{y_1}{6})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-10}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-10-6}{-2-2}\implies \cfrac{-16}{-4}\implies 4 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-6=4(x-2) \\\\\\ y-6=4x-8\implies y=4x-2$

5 0

2 years ago

Read 2 more answers

James had $ 6500 in a bank account that paid 4% interest at the end of each year. How much money did Mr. James have in his accou

Kazeer [188]

He would have $6,760 in his account after year one.
Do 6,500*0.04=260
6,500+260=6,760

8 0

2 years ago

Read 2 more answers

The vertex of a function is (-3, 0) and is a minimum. What is the axis of symmetry and how is it determined?

nalin [4]

The axis of symmetry would be -3 because it is always the (x) of the vertex

5 0

2 years ago

Find the mass of the lamina that occupies the region D = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1} with the density function ρ(x, y) = xye

Alona [7]

Answer:

The mass of the lamina is 1

Step-by-step explanation:

Let $\rho(x,y)$ be a continuous density function of a lamina in the plane region D,then the mass of the lamina is given by:

$m=\int\limits \int\limits_D \rho(x,y) \, dA$ .

From the question, the given density function is $\rho (x,y)=xye^{x+y}$ .

Again, the lamina occupies a rectangular region: D={(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}.

The mass of the lamina can be found by evaluating the double integral:

$I=\int\limits^1_0\int\limits^1_0xye^{x+y}dydx$ .

Since D is a rectangular region, we can apply Fubini's Theorem to get:

$I=\int\limits^1_0(\int\limits^1_0xye^{x+y}dy)dx$ .

Let the inner integral be: $I_0=\int\limits^1_0xye^{x+y}dy$ , then

$I=\int\limits^1_0(I_0)dx$ .

The inner integral is evaluated using integration by parts.

Let $u=xy$ , the partial derivative of u wrt y is

$\implies du=xdy$

and

$dv=\int\limits e^{x+y} dy$ , integrating wrt y, we obtain

$v=\int\limits e^{x+y}$

Recall the integration by parts formula: $\int\limits udv=uv- \int\limits vdu$

This implies that:

$\int\limits xye^{x+y}dy=xye^{x+y}-\int\limits e^{x+y}\cdot xdy$

$\int\limits xye^{x+y}dy=xye^{x+y}-xe^{x+y}$

$I_0=\int\limits^1_0 xye^{x+y}dy$

We substitute the limits of integration and evaluate to get:

$I_0=xe^x$

This implies that:

$I=\int\limits^1_0(xe^x)dx$ .

Or

$I=\int\limits^1_0xe^xdx$ .

We again apply integration by parts formula to get:

$\int\limits xe^xdx=e^x(x-1)$ .

$I=\int\limits^1_0xe^xdx=e^1(1-1)-e^0(0-1)$ .

$I=\int\limits^1_0xe^xdx=0-1(0-1)$ .

$I=\int\limits^1_0xe^xdx=0-1(-1)=1$ .

No unit is given, therefore the mass of the lamina is 1.

3 0

3 years ago