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3 years ago

A students work to simplify a radical is shown below. Select the statemmebt which best applies to rhe sample mathematical work

Mathematics

1 answer:

Alik [6]3 years ago

8 0

Answer:

D. The work shown above is correct, and $y^5\sqrt[4]{y^3}$ may not be simplified further.

Step-by-step explanation:

If you add all the exponents of y in first line then we get

4+4+4+4+4+3=23

Which is same as the exponent in original problem.

So first line is good.

Since it is 4th root so each $y^4$ will turn into y when it moves out of the radical so final answer is correct.

Hence correct choice is "D. The work shown above is correct, and $y^5\sqrt[4]{y^3}$ may not be simplified further." is the correct answer.

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Jim's speedboat can travel 24 miles upstream against a 4 mph current. In the same amount of time it travels 26 miles

8_murik_8 [283]

Answer:

speed of the boat = 100 mph

Step-by-step explanation:

Let x = speed of the boat

upstream speed = x - 4

downstream speed = x + 4

time = 24/(x - 4) = 26/(x + 4)

Cross multiply:

24(x + 4) = 26(x - 4)

24x + 96 = 26x - 104

104 + 96 = 26x - 24x

200 = 2x

x = 100 mph

8 0

2 years ago

Circle O shown below has a radius of 21 inches. Find, to the nearest tenth, the radian

lord [1]

Answer:

He's right it's 0.8 I got the answer right thanks a lot dude

8 0

3 years ago

37. Verify Green's theorem in the plane for f (3x2- 8y2) dx + (4y - 6xy) dy, where C is the boundary of the

Nastasia [14]

I'll only look at (37) here, since

• (38) was addressed in 24438105

• (39) was addressed in 24434477

• (40) and (41) were both addressed in 24434541

In both parts, we're considering the line integral

$\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dy$

and I assume C has a positive orientation in both cases

(a) It looks like the region has the curves y = x and y = x ² as its boundary***, so that the interior of C is the set D given by

$D = \left\{(x,y) \mid 0\le x\le1 \text{ and }x^2\le y\le x\right\}$

• Compute the line integral directly by splitting up C into two component curves,

C₁ : x = t and y = t ² with 0 ≤ t ≤ 1

C₂ : x = 1 - t and y = 1 - t with 0 ≤ t ≤ 1

Then

$\displaystyle \int_C = \int_{C_1} + \int_{C_2} \\\\ = \int_0^1 \left((3t^2-8t^4)+(4t^2-6t^3)(2t))\right)\,\mathrm dt \\+ \int_0^1 \left((-5(1-t)^2)(-1)+(4(1-t)-6(1-t)^2)(-1)\right)\,\mathrm dt \\\\ = \int_0^1 (7-18t+14t^2+8t^3-20t^4)\,\mathrm dt = \boxed{\frac23}$

*** Obviously this interpretation is incorrect if the solution is supposed to be 3/2, so make the appropriate adjustment when you work this out for yourself.

• Compute the same integral using Green's theorem:

$\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dy = \iint_D \frac{\partial(4y-6xy)}{\partial x} - \frac{\partial(3x^2-8y^2)}{\partial y}\,\mathrm dx\,\mathrm dy \\\\ = \int_0^1\int_{x^2}^x 10y\,\mathrm dy\,\mathrm dx = \boxed{\frac23}$

(b) C is the boundary of the region

$D = \left\{(x,y) \mid 0\le x\le 1\text{ and }0\le y\le1-x\right\}$

• Compute the line integral directly, splitting up C into 3 components,

C₁ : x = t and y = 0 with 0 ≤ t ≤ 1

C₂ : x = 1 - t and y = t with 0 ≤ t ≤ 1

C₃ : x = 0 and y = 1 - t with 0 ≤ t ≤ 1

Then

$\displaystyle \int_C = \int_{C_1} + \int_{C_2} + \int_{C_3} \\\\ = \int_0^1 3t^2\,\mathrm dt + \int_0^1 (11t^2+4t-3)\,\mathrm dt + \int_0^1(4t-4)\,\mathrm dt \\\\ = \int_0^1 (14t^2+8t-7)\,\mathrm dt = \boxed{\frac53}$