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3 years ago

PLEASE HELP MEHH!! THANK YOUUUU

Mathematics

2 answers:

ki77a [65]3 years ago

4 0

$\cfrac{3n+5}{7} = \cfrac{3n}{7}+ \cfrac{5}{7}$

Nata [24]3 years ago

4 0

I'm not sure what you want but you can split the terms up into 3n/7 + 5/7.

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The figure below is not drawn to scale. ABCD is a parallelogram, Find

Butoxors [25]

$\angle BAD=100^{\circ}$ (opposite angles in a parallelogram)

$\angle EAD=80^{\circ}$ (subtraction)

$\angle ADE=65^{\circ}$ (angles in a triangle add to 180 degrees)

$\angle CDA=80^{\circ}$ (adjacent angles in a parallelogram are supplementary)

$\boxed{\angle CDE=15^{\circ}}$ (subtraction)

$\angle CED=65^{\circ}$ (angles in a triangle add to 180 degrees)

$\boxed{\angle AEB=80^{\circ}}$ (angles on a straight line add to 180 degrees)

4 0

1 year ago

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

How do I solve that equation

rewona [7]

Solving: (5)(3 - 1) + 4

Step One: Subtract 1 from 3 which is 2
5(3 − 1) + 4

Step Two, Multiply 5 by 2 which is 10=(5)(2) + 4

Step Three, Add 4 to 10 which is 14=10+4

=14

Answer:
(A)14

Hope this helps!

3 0

3 years ago

Read 2 more answers

Help please !! ASAP

Alchen [17]

Answer:

20%

Step-by-step explanation:

6x5=30

20x5=100

7 0

2 years ago

Read 2 more answers

WHICH LINE HAS A SLOPE OF 0 AND GOES THROUGH (5, -8)

arlik [135]

The slope of a horizontal line. A horizontal line has slope 0

3 0

3 years ago