JUST ANSWER RIGHT NOTHING ELSE 25 POINTS!!! plEase thank you

The length of a rectangle is 6 yd longer than its width. if the perimeter of the rectangle is 44 yd , find its area.

NNADVOKAT [17]

Let's call the length L and the width w.

perimeter = 2L + 2w
44 = 2L + 2w

the length is 6 yards longer than the width. so L = w + 6.

plug this into the equation. 44 = 2(w+6) + 2w
44 = 2w + 12 + 2w
44 = 4w + 12
32 = 4w
w = 8
L = w + 6
L = 8 + 6
L = 14

area = Lw
area = 8(14)
area = 112 yd²

3 0

3 years ago

How do I simplify 55/35

Stolb23 [73]

Answer:

1 4/7

Find a number that will divide evenly into both the top number and the bottom number. (You can do any math operation to either number as long as you do the same to the bottom number.)

Keep doing that until you can't find a number that works.

In this case try 5.

55 ÷ 5 = 11 and 35 ÷ 5 = 7

so: 55/35 = 11/7

since the top number is greater than the bottom, reduce 11/7 to 1 4/7

8 0

3 years ago

The ratio of the lengths of the two legs of a right triangle is 3:4

Snowcat [4.5K]

Greetings from Brasil...

Check Attachment!!!

According to the statement of the question, we have:

a/b = 3/4

Note that according to the attached figure, b > a, then the correct proportion will be a to 3, as well as b to 4, because b > a and 4 > 3

Isolating a and b:

a = 3b/4

b = 4a/3

From Pythagoras:

h² = a² + b² - as a = 3b/4 and b = 4a/3, so

For Area:

A = b.a/2

as b = 4h/5 and a = 6h/5, so

4 0

2 years ago

Find the missing side length. and how do i set up to solve?

nikitadnepr [17]

Answer:

its 6 because the lines show they are parellel so it has to be equal to 18 and 12 plus 6 is 18

Step-by-step explanation:

6 0

3 years ago

Find the work required to move an object in the force field F = ex+y <1,1,z> along the straight line from A(0,0,0) to B(-1

storchak [24]

Answer:

Work = e+24

F is not conservative.

Step-by-step explanation:

To find the work required to move an object in the force field

$\large F(x,y,z)=(e^{x+y},e^{x+y},ze^{x+y})$

along the straight line from A(0,0,0) to B(-1,2,-5), we have to parameterize this segment.

Given two points P, Q in any euclidean space, you can always parameterize the segment of line that goes from P to Q with

r(t) = tQ + (1-t)P with 0 ≤ t ≤ 1

so

r(t) = t(-1,2,-5) + (1-t)(0,0,0) = (-t, 2t, -5t) with 0≤ t ≤ 1

is a parameterization of the segment.

the work W required to move an object in the force field F along the straight line from A to B is the line integral

$\large W=\int_{C}Fdr$

where C is the segment that goes from A to B.

$\large \int_{C}Fdr =\int_{0}^{1}F(r(t))\circ r'(t)dt=\int_{0}^{1}F(-t,2t,-5t)\circ (-1,2,-5)dt=\\\\=\int_{0}^{1}(e^t,e^t,-5te^t)\circ (-1,2,-5)dt=\int_{0}^{1}(-e^t+2e^t+25te^t)dt=\\\\\int_{0}^{1}e^tdt-25\int_{0}^{1}te^tdt=(e-1)+25\int_{0}^{1}te^tdt$

Integrating by parts the last integral:

$\large \int_{0}^{1}te^tdt=e-\int_{0}^{1}e^tdt=e-(e-1)=1$

and

$\large \boxed{W=\int_{C}Fdr=e+24}$

To show that F is not conservative, we could find another path D from A to B such that the work to move the particle from A to B along D is different to e+24

Now, let D be the path consisting on the segment that goes from A to (1,0,0) and then the segment from (1,0,0) to B.

The segment that goes from A to (1,0,0) can be parameterized as

r(t) = (t,0,0) with 0≤ t ≤ 1

so the work required to move the particle from A to (1,0,0) is

$\large \int_{0}^{1}(e^t,e^t,0)\circ (1,0,0)dt =\int_{0}^{1}e^tdt=e-1$

The segment that goes from (1,0,0) to B can be parameterized as

r(t) = (1-2t,2t,-5t) with 0≤ t ≤ 1

so the work required to move the particle from (1,0,0) to B is

$\large \int_{0}^{1}(e,e,-5et)\circ (-2,2,-5)dt =25e\int_{0}^{1}tdt=\frac{25e}{2}$

Hence, the work required to move the particle from A to B along D is

e - 1 + (25e)/2 = (27e)/2 -1

since this result differs from e+24, the force field F is not conservative.

6 0

4 years ago