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2 years ago

If a rat grows 1.5 inches a year, how many years will it take the rat to grow to 12.3 inches?

Mathematics

2 answers:

vladimir1956 [14]2 years ago

7 0

You would do 12.3/1.5= 8.2 years

Vikki [24]2 years ago

3 0

You divide 12.3 by 1.5 and you'll get 8.2
i hope you under stand how to solve this problem

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5*9 (sqrt 10)^3= Help please

Nikitich [7]

Use BODMAS or BIDMAS (Brackets, over Indices, Division, Multiplication, Addition, the Subtraction, this tells you which order to do things in)

As you cant do √10 in the brackets you do the indices, so (√10)³
Split this up to make it easier
(√10)³= √10 x √10 x √10 = 10√10

You the multiply this by 9

9 x 10√10 = 90√10

then multiply by 5

5 x 90√10 = 450√10 = 1423.024947 (using calculator)

7 0

2 years ago

Given WRST is a parallelogram and

daser333 [38]

It would be a parallelogram

3 0

3 years ago

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

Two perpendicular lines intersect at the origin. If the slope of the first line is .5, what is the equation of the second line?.

Sedaia [141]

The origin is at the point (0,0). Therefore, the y-intercept will be 0. For two points to be perpendicular, the slopes must be opposite reciprocals of each other. .5 can be seen as 1/2, so the reciprocal is 2, and it is positive, making that 2 negative. Your equation would be y = -2x.

3 0

3 years ago

Please help me thanks

KiRa [710]

Answer:

1,2,4,5,10

1,2,3,5,6

Step-by-step explanation:

7 0

3 years ago