What is the perimeter of the triangle? khan

Can u guys please give me the correct answer

Andre45 [30]

Answer:

∠ 3 = 65°

Step-by-step explanation:

∠ 2 and 115° are a linear pair and sum to 180° , that is

∠ 2 + 115° = 180° ( subtract 115° from both sides )

∠ 2 = 65°

∠ 2 and ∠ 3 are corresponding angles and and are congruent , then

∠ 3 = 65°

6 0

2 years ago

The graph below shows a company's profit f(x), in dollars, depending on the price of pencils x, in dollars, sold by the company:

jarptica [38.1K]

Answer:

the constraints of the domain are negative and loose

Step-by-step explanation:

8 0

3 years ago

Find the equation (in slope-intercept form) of the line passing through the points with the given coordinates. (3, −4) (4,-5)

leva [86]

Y=-1x-1 is your formula.

5 0

4 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

Can I get some help? Thanks!

vivado [14]

Answer:

C: y=11x

Step-by-step explanation:

11x2=22

11x3=33

11x4=44

So on and so forth.

4 0

3 years ago

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