In the fraction 15 over 7 which of the following is 15

Use the divergence theorem to find the outward flux of the vector field F(xyz)= 4x^2 i + 4y^2 j + 3z^2 k across the boundary of

Pavel [41]

$\vec F(x,y,z)=4x^2\,\vec\imath+4y^2\,\vec\jmath+3z^2\,\vec k\implies\nabla\cdot\vec F=8x+8y+6z$

Let $S$ be the surface of the rectangular prism bounded by the planes $x=0$ , $x=a$ , $y=0$ , $y=b$ , $z=0$ , and $z=c$ . By the divergence theorem, the integral of $\vec F$ over $S$ is given by the integral of $\nabla\cdot\vec F$ over the interior of $S$ (call it $R$ ):

$\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\iiint_R(\nabla\cdot\vec F)\,\mathrm dV$

$=\displaystyle\int_0^c\int_0^b\int_0^a(8x+8y+6z)\,\mathrm dx\,\mathrm dy\,\mathrm dz=\boxed{abc(4a+4b+3c)}$

5 0

3 years ago

A figure is composed of a rectangle and a right triangle. What is the area of the figure?

strojnjashka [21]

QUESTION 1

The figure is a tra-pezoid.

The area of a trapezoid is given by the formula;

$A=\frac{1}{2}(Sum\:of\:parallel\:sides)\times h$

We substitute the values to obtain;

$Area=\frac{1}{2}(13.2cm+8.4cm)\times 6cm$

$Area=\frac{1}{2}(21.6cm)\times 6cm$

$Area=64.8cm^2$

QUESTION 2

The approximate length of the ribbon is equal to the circumference of the circular table cloth.

We can find this by using the formula for calculating the circumference of a circle.

$C=2\pi r$

where $r=3.5ft$ is the radius of the circular table cloth.

We substitute this value and $\pi=3.14$ into the formula to get;

$C=2(3.14)(3.5)ft$

$C=21.98ft$

The approximate length of the ribbon is 22.0ft to the nearest tenth.

QUESTION 3

Type of quadrilateral:Rectangle

Explanation: A rectangle has 4 angles that are right angles.

The two pairs of opposite sides of a rectangle are also congruent.

QUESTION 4.

The circumference of a semi circle is calculated using the formula;

$C=\pi r$

where $r=12.5cm$ is the radius of the semicircle.

$C=3.14\times 12.5cm$

$C=39.25cm$

QUESTION 5

The area of the entire figure is the area of the semicircle plus the area of the isosceles triangle.

$Area=\frac{1}{2}\pi r^2+\frac{1}{2}bh$

$Area=\frac{1}{2}\times3.14\times12.5^2+\frac{1}{2}\times25\times 24$

$Area=345.3125+300$

$Area=645.3125cm^2$

$Area\approx645cm^2$

QUESTION 6

The area of the parallelogram is given by the formula;

$A=bh$

From the diagram, $b=y\:in.,h=3in.$ .

Given, A=23.7 inches squared.

We substitute the values to obtain;

$23.7=3y$

$y=\frac{23.7}{3}$

$y=7.9in.$

QUESTION 7

The area of a rectangle is given by the formula;

$Area=l\times b$

The area of the bigger rectangle $=9\times15=135in^2$

The area of the smaller rectangle $=8\times 13=104in^2$

The area of the shaded region $=135-104=31in^2$

5 0

3 years ago

20. Acid Mixture Bill Charlson needs 10% hydrochloric acid

nevsk [136]

5% more to the mixture

5 0

3 years ago

Read 2 more answers

Find f(x) if it is known that f(5a)=5a-2

Romashka-Z-Leto [24]

Answer:

f(x)=x-2

Step-by-step explanation:

You just have to change 5a for x

8 0

2 years ago

Conditional Probability

trapecia [35]

To solve for P(QIR) we use the formula:
P(QIR)=[P(Q∩R)]/P(R)
But from the diagram:
P(Q∩R)=3/22
P(R)=7/22
hence
P(QIR)=3/22÷7/22=3/7

3 0

3 years ago