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3 years ago

Beth rides 1 7/32 kristine rides 39/32 who rides the farthest

1 answer:

Citrus2011 [14]3 years ago

3 0

Neither, they ride the same.

First, convert 39/32 to a mixed number. Since it equals 1 7/32, and that's how much beth rode, they both rode the same.

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What is 6x^3+20x^2+15x+3 divided by 3x+1

torisob [31]

Answer:

2x^2+6x+3

Step-by-step explanation:

answer : 2x^2+6x+3

3 0

3 years ago

Pease help, thank you

andreyandreev [35.5K]

Answer: -2

=====================================================

Draw a vertical line through 4 on the x axis. This vertical line crosses the parabola at some point (which we'll call point A). Draw a horizontal line from point A to the y axis and note how it lands on y = 12. Therefore the point (4,12) is on this parabola.

Repeat the same steps as before to find that (8,4) is also on the parabola

We need to find the slope of the line through (4,12) and (8,4)

m = (y2 - y1)/(x2 - x1)

m = (4-12)/(8 - 4)

m = -8/4

m = -2

The slope of this line is -2 meaning that the average rate of change from x = 4 to x = 8 is -2.

The line goes down 2 units each time you move to the right 1 unit.

6 0

3 years ago

Ash buys 5 bunches of flowers. Each bunch contains 10 flowers. He gets 8 more flowers from his yard.

Semenov [28]

Answer:

5 x 10 + 8 = f

Step-by-step explanation:

The reason it is 5 x 10 + 8 = f, is because each bunch has 10 and Ash has five.

In order to find how many flowers he has in 5 bunches is to multiply by how many flowers each bunch has.

Then he gets 8 more flowers which is why it has +8 in the equation.

6 0

2 years ago

Read 2 more answers

Length = 2 feet, width = 15 inches, height = 8 inches<br><br><br><br> V= _____________

I am Lyosha [343]

Volume = length * width ( aka base ) * height

2 feet = 24 inches.

24*15*8=2880 inches.

6 0

3 years ago

Read 2 more answers

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S.

sweet-ann [11.9K]

Answer:

2.794

Step-by-step explanation:

Recall that if G(x,y) is a parametrization of the surface S and F and G are smooth enough then

$\bf \displaystyle\iint_{S}FdS=\displaystyle\iint_{R}F(G(x,y))\cdot(\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y})dxdy$

F can be written as

F(x,y,z) = (xy, yz, zx)

and S has a straightforward parametrization as

$\bf G(x,y) = (x, y, 3-x^2-y^2)$

with 0≤ x≤1 and 0≤ y≤1

$\bf \displaystyle\frac{\partial G}{\partial x}= (1,0,-2x)\\\\\displaystyle\frac{\partial G}{\partial y}= (0,1,-2y)\\\\\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y}=(2x,2y,1)$

we also have

$\bf F(G(x,y))=F(x, y, 3-x^2-y^2)=(xy,y(3-x^2-y^2),x(3-x^2-y^2))=\\\\=(xy,3y-x^2y-y^3,3x-x^3-xy^2)$

and so

$\bf F(G(x,y))\cdot(\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y})=(xy,3y-x^2y-y^3,3x-x^3-xy^2)\cdot(2x,2y,1)=\\\\=2x^2y+6y^2-2x^2y^2-2y^4+3x-x^3-xy^2$

we just have then to compute a double integral of a polynomial on the unit square 0≤ x≤1 and 0≤ y≤1

$\bf \displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1}(2x^2y+6y^2-2x^2y^2-2y^4+3x-x^3-xy^2)dxdy=\\\\=2\displaystyle\int_{0}^{1}x^2dx\displaystyle\int_{0}^{1}ydy+6\displaystyle\int_{0}^{1}dx\displaystyle\int_{0}^{1}y^2dy-2\displaystyle\int_{0}^{1}x^2dx\displaystyle\int_{0}^{1}y^2dy-2\displaystyle\int_{0}^{1}dx\displaystyle\int_{0}^{1}y^4dy+\\\\+3\displaystyle\int_{0}^{1}xdx\displaystyle\int_{0}^{1}dy-\displaystyle\int_{0}^{1}x^3dx\displaystyle\int_{0}^{1}dy-\displaystyle\int_{0}^{1}xdx\displaystyle\int_{0}^{1}y^2dy$

=1/3+2-2/9-2/5+3/2-1/4-1/6 = 2.794

5 0

3 years ago