Need help What is 50 x 10 to the 2nd power in standard form?

Let R = [ 0 , 1 ] × [ 0 , 1 ] R=[0,1]×[0,1]. Find the volume of the region above R R and below the plane which passes through th

Bond [772]

The three vectors $\langle0,0,1\rangle$ , $\langle1,0,8\rangle$ , and $\langle0,1,9\rangle$ each terminate on the plane. We can get two vectors that lie on the plane itself (or rather, point in the same direction as vectors that do lie on the plane) by taking the vector difference of any two of these. For instance,

$\langle1,0,8\rangle-\langle0,0,1\rangle=\langle1,0,7\rangle$

$\langle0,1,9\rangle-\langle0,0,1\rangle=\langle0,1,8\rangle$

Then the cross product of these two results is normal to the plane:

$\langle1,0,7\rangle\times\langle0,1,8\rangle=\langle-7,-8,1\rangle$

Let $(x,y,z)$ be a point on the plane. Then the vector connecting $(x,y,z)$ to a known point on the plane, say (0, 0, 1), is orthogonal to the normal vector above, so that

$\langle-7,-8,1\rangle\cdot(\langle x,y,z\rangle-\langle0,0,1\rangle)=0$

which reduces to the equation of the plane,

$-7x-8y+z-1=0\implies z=7x+8y+1$

Let $z=f(x,y)$ . Then the volume of the region above $R$ and below the plane is

$\displaystyle\int_0^1\int_0^1(7x+8y+1)\,\mathrm dx\,\mathrm dy=\boxed{\frac{17}2}$

6 0

2 years ago

A guitarist sets a goal to practice 200 minutes each week. If he only is going to practice 5 days during the week, which of the

vodomira [7]

Answer:

40

Step-by-step explanation:

40x5=200

200 divided by 5 =40

5 0

2 years ago

Which function f (x) , graphed below, or g (x) , whose equation is g (x) = 3 cos 1/4 (x + x/3) + 2, has the largest maximum and

Leya [2.2K]

Answer:

g(x), and the maximum is 5

Step-by-step explanation:

for given function f(x), the maximum can be seen from the shown graph i.e. 2

But for the function g(x), maximum needs to be calculated.

Given function :

g (x) = 3 cos 1/4 (x + x/3) + 2

let x=0 (as cosine is a periodic function and has maximum value of 1 at 0 angle)

g(x)= 3 cos1/4(0 + 0) +2

= 3cos0 +2

= 3(1) +2

= 3 +2

= 5 !

7 0

2 years ago

Find the exact length of the curve. x=et+e−t, y=5−2t, 0≤t≤2 For a curve given by parametric equations x=f(t) and y=g(t), arc len

Rama09 [41]

The length of a curve C parameterized by a vector function r(t) = x(t) i + y(t) j over an interval a ≤ t ≤ b is

$\displaystyle\int_C\mathrm ds = \int_a^b \sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2} \,\mathrm dt$

In this case, we have

x(t) = exp(t ) + exp(-t ) ==> dx/dt = exp(t ) - exp(-t )

y(t) = 5 - 2t ==> dy/dt = -2

and [a, b] = [0, 2]. The length of the curve is then

$\displaystyle\int_0^2 \sqrt{\left(e^t-e^{-t}\right)^2+(-2)^2} \,\mathrm dt = \int_0^2 \sqrt{e^{2t}-2+e^{-2t}+4}\,\mathrm dt$

$=\displaystyle\int_0^2 \sqrt{e^{2t}+2+e^{-2t}} \,\mathrm dt$

$=\displaystyle\int_0^2\sqrt{\left(e^t+e^{-t}\right)^2} \,\mathrm dt$

$=\displaystyle\int_0^2\left(e^t+e^{-t}\right)\,\mathrm dt$

$=\left(e^t-e^{-t}\right)\bigg|_0^2 = \left(e^2-e^{-2}\right)-\left(e^0-e^{-0}\right) = \boxed{e^2-\frac1{e^2}}$

5 0

2 years ago

Look at the figure which step should be taken next to construct a line parallel to AB

FinnZ [79.3K]

My answer is the 2nd option.

Without changing the compass setting from the previous step, place the compass on point P. Draw an arc similar to the one already drawn.

Parallel lines are lines that do not meet. In this figure, point P is the point where the 2nd line can be drawn and become parallel to line AB.

8 0

3 years ago