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3 years ago

10

Give two benefits of knowing what is in your credit report, not your credit score. (Use at least one complete sentence for each

benefit.)

1 answer:

seraphim [82]3 years ago

7 0

Gvvv. Dbbznxbdncbndbzbsbxbdbdjdbdbdbdbdjdbdjd

You might be interested in

Use Stokes' Theorem to evaluate S curl F · dS. F(x, y, z) = x2 sin(z)i + y2j + xyk, S is the part of the paraboloid z = 9 − x2 −

Korolek [52]

The vector field

$\vec F(x,y,z)=x^2\sin z\,\vec\imath+y^2\,\vec\jmath+xy\,\vec k$

has curl

$\nabla\times\vec F(x,y,z)=x\,\vec\imath+(x^2\cos z-y)\,\vec\jmath$

Parameterize $S$ by

$\vec s(u,v)=x(u,v)\,\vec\imath+y(u,v)\,\vec\jmath+z(u,v)\,\vec k$

where

$\begin{cases}x(u,v)=u\cos v\\y(u,v)=u\sin v\\z(u,v)=(9-u^2)\end{cases}$

with $0\le u\le3$ and $0\le v\le2\pi$ .

Take the normal vector to $S$ to be

$\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}=2u^2\cos v\,\vec\imath+2u^2\sin v\,\vec\jmath+u\,\vec k$

Then by Stokes' theorem we have

$\displaystyle\int_{\partial S}\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{2\pi}\int_0^3(\nabla\times\vec F)(\vec s(u,v))\cdot\left(\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{2\pi}\int_0^3u^3(2u\cos^3v\sin(u^2-9)+\cos^3v\sin v+2u\sin^3v+\cos v\sin^3v)\,\mathrm du\,\mathrm dv$

which has a value of 0, since each component integral is 0:

$\displaystyle\int_0^{2\pi}\cos^3v\,\mathrm dv=0$

$\displaystyle\int_0^{2\pi}\sin v\cos^3v\,\mathrm dv=0$

$\displaystyle\int_0^{2\pi}\sin^3v\,\mathrm dv=0$

$\displaystyle\int_0^{2\pi}\cos v\sin^3v\,\mathrm dv=0$

4 0

3 years ago

Full steps and all reasons, thanks!

Nataly_w [17]

You can focus on the two sub-triangles ABC and ACD

We know that they are both right triangles, because we're given that AC and BD are perpendicular.

The sides AB and AD are the hypothenuses of these triangles, so we know their length through the pythagorean theorem:

$AD^2 = AC^2+CD^2$

$AB^2 = AC^2+BC^2$

But since we're given $BC=CD$ , the expressions for AD and AB are actually the same expression. This proves that AD=AB, and thus that the triangle is isosceles.

8 0

4 years ago

How do I solve 5x - 4 = -19

ira [324]

Undo what has been done to the variable. That means to do the opposite done to the variable.
5x-4=-19
+4=+4
5x=-15
5 is being multiplied to the variable so you divide it from both sides.
x=-3

7 0

3 years ago

Nana76 [90]

Answer:

no I do not agree, Because it doesn't make since

6 0

2 years ago

H and J are both between G and K such that GH = JK and H is between G and J.

DIA [1.3K]

Answer:

32

Step-by-step explanation:

So the order of the points is: GHJK

GH = x+10 =JK

HJ=8

JK=2x-4

JK = x+10=2x-4

x+10=2x-4

10+4=2x-x

x=14

GH= 14+10 =24

GJ = GH +HJ = 24+ 8=32

6 0

3 years ago