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3 years ago

Help please ;-;

Mathematics

1 answer:

anzhelika [568]3 years ago

3 0

Answer:

could be either b or d

Step-by-step explanation:

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Will someone please help me here

frez [133]

Hey!

You have to divide the total amount she has by the surface area.

Surface area = 2(wl+hl+hw)

That gives you:

$(15 + 18 + 30) \times 2 \\ 63 \times 2 = 126$

The surface area is 126 ft. Now divide the total amount she has by the surface area.

$1,134 \div 126 = 9$

<em>That means she can cover 9 boxes</em>

$\framebox{She can cover 9 boxes}$

8 0

3 years ago

PLEASE PLEASE PLEASE HELP ME

qaws [65]

Answer:

$\large\boxed{\dfrac{1}{45}}$

Step-by-step explanation:

2 yellow marables of 10 all marables

$P(A_1)=\dfrac{2}{10}=\dfrac{1}{5}$

1 yellow marable of 9 all marables

$P(A_2)=\dfrac{1}{9}$

$P(A)=P(A_1)\cdot P(A_2)\\\\P(A)=\dfrac{1}{5}\cdot\dfrac{1}{9}=\dfrac{1}{45}$

5 0

3 years ago

Find the slope given 2 points: (-2, 1) and (-1,7)

Strike441 [17]

Answer:

- 1,7

Step-by-step explanation:

but not sure on this if wrong so sorry

7 0

3 years ago

The proof that is shown. Given: ΔMNQ is isosceles with base , and and bisect each other at S. Prove: Square M N Q R is shown wit

Afina-wow [57]

Answer:

MS and QS

Step-by-step explanation:

Finished the test!!! Hope u have a great day or night wherever u are :)

5 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