All categories

3 years ago

Carpeting costs 9.99 a yard. If jan buys 17.4 yards, how much will it cost her (round your anwser to the nearest hundreth

Mathematics

1 answer:

aalyn [17]3 years ago

4 0

To get the answer, multiply 9.99 x 17.4 yards. That will give you the answer 173.826. That rounds to 173.83

You might be interested in

A portable CD player costs $48. How much is 25% of the cost

blsea [12.9K]

$x = 48 \times .25$
x equals 25% of the cost.
.25 equals 25%
Therefore x=12
The awnser is $12

7 0

2 years ago

Simplify -5(20 + 3).

sammy [17]

Whenever a number is beside a parenthesis without a sign it means multiplication. So you would use PEMDAS work in parenthesis then exponents, multiplication (from left to right) addition and subtraction (from left to right). so you would add 20+3 and get 23 then multiply that by -5 to get -115 as your answer.

4 0

3 years ago

Read 2 more answers

HELP ILL GIVE YOU BRAINLIEST ‼️‼️‼️‼️

Sladkaya [172]

Answer:

36 200 a-c = 36 for 200 and that is how it's done

5 0

3 years ago

one evening 1500 concert tickets were sold for the Jazz Fest. Tickets cost $30 for covered pavillion seats and $10 for lawn seat

snow_lady [41]

$x-number\ of\ tickets\ for\ pavilion\\1500-x-number\ of\ tickets\ for\ lawn\\\\30x+10(1500-x)=27000\\30x+15000-10x=27000\\30x-10x=27000-15000\\20x=12000\ \ \ \ \ \ \ \ |:20\\\\x=600\\1500-600=900$

6 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

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

• Using Green's theorem:

$\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dx = \int_0^1\int_0^{1-x}10y\,\mathrm dy\,\mathrm dx = \boxed{\frac53}$

4 0

3 years ago