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2 years ago

6

On July 3, the ABC corporation closed at $43.67. This was a $0.47 increase from the close the day before. What percentage change

is this?

1 answer:

Digiron [165]2 years ago

3 0

$0.47 is about 1.076%
Divide $0.47 by $43.67 to get your answer

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What is 1/4 of 76 in a mathequ on

Softa [21]

The answer is 19. You have to do 1/4 x 76/1 which is basically 76 divided by 4, which gives you 19.

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3 years ago

Which ordered pair is a solution of the system of a linear equation below? Y= 1/4x +2 y= x-1

Tanya [424]

Answer: x = 4, y = 3

<u>Step-by-step explanation:</u>

$y=\dfrac{1}{4}x+2$ and $y=x-1$

Using substitution method we get:

$x-1=\dfrac{1}{4}x+2\\\\\text{subtract}\ \dfrac{1}{4}x\ \text{from both sides}\rightarrow \dfrac{3}{4}x-1=2\\\\\text{add 1 to both sides}\rightarrow \dfrac{3}{4}x=3\\\\\text{multiply both sides by 4}\rightarrow 3x=12\\\\\text{divide both sides by 3}\rightarrow x=4\\\\\text{Insert x = 4 into one of the original equations to solve for y:}\\y=x-1\\y =(4)-1\\y=3$

3 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 <em>C</em> has a positive orientation in both cases

(a) It looks like the region has the curves <em>y</em> = <em>x</em> and <em>y</em> = <em>x</em> ² as its boundary***, so that the interior of <em>C</em> is the set <em>D</em> 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 <em>C</em> into two component curves,

<em>C₁ </em>: <em>x</em> = <em>t</em> and <em>y</em> = <em>t</em> ² with 0 ≤ <em>t</em> ≤ 1

<em>C₂</em> : <em>x</em> = 1 - <em>t</em> and <em>y</em> = 1 - <em>t</em> with 0 ≤ <em>t</em> ≤ 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) <em>C</em> 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 <em>C</em> into 3 components,

<em>C₁</em> : <em>x</em> = <em>t</em> and <em>y</em> = 0 with 0 ≤ <em>t</em> ≤ 1

<em>C₂</em> : <em>x</em> = 1 - <em>t</em> and <em>y</em> = <em>t</em> with 0 ≤ <em>t</em> ≤ 1

<em>C₃</em> : <em>x</em> = 0 and <em>y</em> = 1 - <em>t</em> with 0 ≤ <em>t</em> ≤ 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

the area of the top of a rectangular table is 323 square feet. If the length of the table is 19 feet , what is the width

sergejj [24]

Answer:

hello :

the area is : A= w ×l ... w : the width l : the length

A = 323 l = 19 w ?

l = A/w

l = 323/ 19

l = 17 feet

4 0

3 years ago

Find the endpoint:

scZoUnD [109]

Answer:

The coordinates of Point D are: (-5,-8)

Step-by-step explanation:

The formula for mid-point is given by:

$M = (\frac{x_1+x_2}{2} , \frac{y_1+y_2}{2})$

For the point (x1,y1) and (x2,y2)

Given that

M = (-3, -6)

C = (-1, -4) => (x1,y1)

Putting the values of given points in the formula for mid-point

$(-3,-6) = (\frac{-1+x_2}{2} , \frac{-4+y_2}{2})$

Putting respective coordinates equal

$-3 = \frac{-1+x_2}{2}\\-6 =-1+x_2\\x_2=-6+1\\x_2 = -5\\AND\\-6 = \frac{-4+y_2}{2}\\-12 = -4+y_2\\y_2 = -12+4\\y_2 = -8$

Hence,

The coordinates of Point D are: (-5,-8)

3 0

2 years ago