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3 years ago

The solid figure is separated along the dotted line into two rectangular prisms.

Mathematics

2 answers:

ExtremeBDS [4]3 years ago

7 0

54 ??? Should be the answer

Alla [95]3 years ago

3 0

Answer:54

Step-by-step explanation:

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3. Which inequality matches the graph ? *

luda_lava [24]

X is greater than or equal to 6, this means it’s choice C.

4 0

3 years ago

Amelia and Joey decided to shoot arrows at a simple target with a large outer ring and a smaller bull's-eye. Amelia went first a

Alona [7]

Answer:

outer ring worth 14 pts

bull's-eye worth 74.333333 pts

Step-by-step explanation:

let the worth point of landing an arrow on the outer ring be "x" and on bull's eye be "y"

For amelia

$3x + 3y = 267$

For joey

$4x + 3y = 281$

subtracting the first equation from the second

$x = 14$

$3x + 3y = 267$

$42 + 3y = 267$

$3y = 223$

$y = 74.333$

3 0

4 years ago

Rodney is starting a tutoring business.

sweet-ann [11.9K]

Answer:

(a) 20 + 14(s)

(b)$146

Step-by-step explanation:

Use s as your variable per session.

$20 is a flat fee as a sign up for tutoring.

Therefore, 20 + 14s is the expression.

(b) 14x9=126 + 20 = 146

7 0

2 years ago

*4.1.9

Cerrena [4.2K]

The answer would be:
7 to 11
7;11
7/11

8 0

3 years ago

Find the curl and the divergence of the vector field f(x, y, z) = x2yz i + xy2z j + xyz2 k

xxTIMURxx [149]

$\mathbf f(x,y,z)=x^2yz\,\mathbf i+xy^2z\,\mathbf j+xyz^2\,\mathbf k$

$\mathrm{div}\mathbf f(x,y,z)=\dfrac{\partial(x^2yz)}{\partial x}+\dfrac{\partial(xy^2z)}{\partial y}+\dfrac{\partial(xyz^2)}{\partial z}$
$=2xyz+2xyz+2xyz$
$=6xyz$

$\mathrm{curl}\mathbf f(x,y,z)=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\\\\frac\partial{\partial x}&\frac\partial{\partial y}&\frac\partial{\partial z}\\\\x^2yz&xy^2z&xyz^2\end{vmatrix}$
$=\left(\dfrac{\partial(xyz^2)}{\partial y}-\dfrac{\partial(xy^2z)}{\partial z}\right)\,\mathbf i-\left(\dfrac{\partial(xyz^2)}{\partial x}-\dfrac{\partial(x^2yz)}{\partial z}\right)\,\mathbf j+\left(\dfrac{\partial(xy^2z)}{\partial x}-\dfrac{\partial(x^2yz)}{\partial y}\right)\,\mathbf k$
$=\left(xz^2-xy^2\right)\,\mathbf i-\left(yz^2-x^2y\right)\,\mathbf j+\left(y^2z-x^2z\right)\,\mathbf k$

6 0

4 years ago