Find the jacobian of the transformation. x = 6v + 6w2, y = 8w + 8u2, z = 2u + 2v2

Each locker is shaped like a rectangular prism. Which has more storage space? explain/

seropon [69]

Locker 1 has more storage space because it has a greater volume.
The volume of locker 1 is 8,640 (48x15x12)
The volume of locker 2 is 7,200 (60x10x12)

I hope this helps.

4 0

2 years ago

A volunteer is organizing the snacks at a concession stand and notices the ratio is sweet snacks to salty snacks is 5:1. If ther

Verizon [17]

100 sweet snack, if it is 5:1 ratio than there’s 5 times the amount of salty snacks

3 0

2 years ago

Evaluate the triple integral ∭ExydV where E is the solid tetrahedon with vertices (0,0,0),(5,0,0),(0,9,0),(0,0,4).

Elan Coil [88]

Answer: $\int\limits^a_E {\int\limits^a_E {\int\limits^a_E {xy} } \, dV$ = 1087.5

Step-by-step explanation: To evaluate the triple integral, first an equation of a plane is needed, since the tetrahedon is a geometric form that occupies a 3 dimensional plane. The region of the integral is in the attachment.

An equation of a plane is found with a point and a normal vector. Normal vector is a perpendicular vector on the plane.

Given the points, determine the vectors:

P = (5,0,0); Q = (0,9,0); R = (0,0,4)

vector PQ = (5,0,0) - (0,9,0) = (5,-9,0)

vector QR = (0,9,0) - (0,0,4) = (0,9,-4)

Knowing that cross product of two vectors will be perpendicular to these vectors, you can use the cross product as normal vector:

n = PQ × QR = $\left[\begin{array}{ccc}i&j&k\\5&-9&0\\0&9&-4\end{array}\right]$ $\left[\begin{array}{ccc}i&j\\5&-9\\0&9\end{array}\right]$

n = 36i + 0j + 45k - (0k + 0i - 20j)

n = 36i + 20j + 45k

Equation of a plane is generally given by:

$a(x-x_{0}) + b(y-y_{0}) + c(z-z_{0}) = 0$

Then, replacing with point P and normal vector n:

$36(x-5) + 20(y-0) + 45(z-0) = 0$

The equation is: 36x + 20y + 45z - 180 = 0

Second, in evaluating the triple integral, set limits:

In terms of z:

$z = \frac{180-36x-20y}{45}$

When z = 0:

$y = 9 + \frac{-9x}{5}$

When z=0 and y=0:

x = 5

Then, triple integral is:

$\int\limits^5_0 {\int\limits {\int\ {xy} \, dz } \, dy } \, dx$

Calculating:

$\int\limits^5_0 {\int\limits {\int\ {xyz} \, dy } \, dx$

$\int\limits^5_0 {\int\limits {\int\ {xy(\frac{180-36x-20y}{45} - 0 )} \, dy } \, dx$

$\frac{1}{45} \int\limits^5_0 {\int\ {180xy-36x^{2}y-20xy^{2}} \, dy } \, dx$

$\frac{1}{45} \int\limits^5_0 {90xy^{2}-18x^{2}y^{2}-\frac{20}{3} xy^{3} } \, dx$

$\frac{1}{45} \int\limits^5_0 {2430x-1458x^{2}+\frac{94770}{125} x^{3}-\frac{23490}{375}x^{4} } \, dx$

$\frac{1}{45} [30375-60750+118462.5-39150]$

$\int\limits^5_0 {\int\limits {\int\ {xyz} \, dy } \, dx$ = 1087.5

The volume of the tetrahedon is 1087.5 cubic units.

3 0

1 year ago

The first term of a finite geometric series is 6 and the last term is 4374. The sum of all the term os 6558. find the common rat

Oliga [24]

A geometric series is written as $ar^n$ , where $a$ is the first term of the series and $r$ is the common ratio.

In other words, to compute the next term in the series you have to multiply the previous one by $r$ .

Since we know that the first time is 6 (but we don't know the common ratio), the first terms are

$6, 6r, 6r^2, 6r^3, 6r^4, 6r^5, \ldots$ .

Let's use the other information, since the last term is $4374 > 6$ , we know that $r>1$ , otherwise the terms would be bigger and bigger.

The information about the sum tells us that

$\displaystyle \sum_{i=0}^n 6r^i = 6\sum_{i=0}^n r^i = 6558$

We have a formula to compute the sum of the powers of a certain variable, namely

$\displaystyle \sum_{i=0}^n r^i = \cfrac{r^{n+1}-1}{r-1}$

So, the equation becomes

$6\cfrac{r^{n+1}-1}{r-1} = 6558$

The only integer solution to this expression is $n=6, r=3$ .

If you want to check the result, we have

$6+6*3+6*3^2+6*3^3+6*3^4+6*3^5+6*3^6 = 6558$

and the last term is

$6*3^6 = 4374$

7 0

2 years ago

The diagram shows a logo made from three circles

ratelena [41]

Answer:

Use the formula π*(r^2) where r is radius

Area of big circle, all 3=314.1592654 (approximately) and this is =100%

Area of middle circle=153.93804

Area of small circle=78.53981634

Percentage of middle circle with small circle = (153.93804/314.1592654)*100= 48.999999999999 approx= 49%

Percentage of small circle alone = (78.53981634/314.1592654)*100

= 25%

So 51%= big circle alone

And 51%+25%= 76%

100%-76%=24%

3 0

2 years ago

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