PLEASE HELP

The weight of 3 fruit baskets are shown below. Basket A 12 Kg / Basket B 8 Kg / Basket C 18 kg Which basket is heaviest?

Feliz [49]

I think it is B I hope I am right and helped

5 0

3 years ago

I need help with this geometry question.

sashaice [31]

Answer:

Step-by-step explanation:

a). There are 8 points in the figure attached.

b). There are 9 lines in the given figure.

c). There are 5 planes in the figure attached.

d). Three collinear points are D,G and F.

e). Four co-planar points are G, F, H and C.

f). Intersection of planes ABC and ABE is the common line AB.

g). Intersection of planes BCH and DEF is the common line EF.

h). Intersection of AD and DF is a point D.

3 0

3 years ago

Quadrilateral WXYZ is on a coordinate plane. Segment XY is on the line x − 3y = −12, and segment WZ is on the x − 3y = −6. Which

Nostrana [21]

Answer:

Equation shown

Step-by-step explanation:

Eq(1): x-3y=-12

X=-12+3y

=-4+y

Eq(2): x-3y=-6

X=-6+3y

=-2+y

4 0

3 years ago

Read 2 more answers

Evaluate the triple integral ∭EzdV where E is the solid bounded by the cylinder y2+z2=81 and the planes x=0,y=9x and z=0 in the

dem82 [27]

Answer:

I = 91.125

Step-by-step explanation:

Given that:

$I = \int \int_E \int zdV$ where E is bounded by the cylinder $y^2 + z^2 = 81$ and the planes x = 0 , y = 9x and z = 0 in the first octant.

The initial activity to carry out is to determine the limits of the region

since curve z = 0 and $y^2 + z^2 = 81$

∴ $z^2 = 81 - y^2$

$z = \sqrt{81 - y^2}$

Thus, z lies between 0 to $\sqrt{81 - y^2}$

GIven curve x = 0 and y = 9x

$x =\dfrac{y}{9}$

As such,x lies between 0 to $\dfrac{y}{9}$

Given curve x = 0 , $x =\dfrac{y}{9}$ and z = 0, $y^2 + z^2 = 81$

y = 0 and

$y^2 = 81 \\ \\ y = \sqrt{81} \\ \\ y = 9$

∴ y lies between 0 and 9

Then $I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \int^{\sqrt{81-y^2}}_{z=0} \ zdzdxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{z^2}{2} \end {bmatrix} ^ {\sqrt {{81-y^2}}}_{0} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{(\sqrt{81 -y^2})^2 }{2}-0 \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{{81 -y^2} }{2} \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81x -xy^2} }{2} \end {bmatrix} ^{\dfrac{y}{9}}_{0} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81(\dfrac{y}{9}) -(\dfrac{y}{9})y^2} }{2}-0 \end {bmatrix} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81 \ y -y^3} }{18} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \int^9_{y=0} \begin {bmatrix} {81 \ y -y^3} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \begin {bmatrix} {81 \ \dfrac{y^2}{2} - \dfrac{y^4}{4}} \end {bmatrix}^9_0$

$I = \dfrac{1}{18} \begin {bmatrix} {40.5 \ (9^2) - \dfrac{9^4}{4}} \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 3280.5 - 1640.25 \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 1640.25 \end {bmatrix}$

I = 91.125

4 0

3 years ago

More Algebra hw-- :/

grin007 [14]

Answer:
-11x=-66
x=6

Explanation:
-11x-12=-78
1) Add 12 to both sides.
-11x=-78+12
-11x=-66
2) Divide both sides by -11.
x=-66/-11
x=6

Hope this helps! Please leave a thanks and brainliest answer if you can :)

3 0

2 years ago