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2 years ago

Please help with math 4r+2+7r

Mathematics

2 answers:

olga nikolaevna [1]2 years ago

6 0

The simplified expression is 11r+2

UkoKoshka [18]2 years ago

6 0

Answer:

11r + 2

Step-by-step explanation:

I combined liked terms which are the r's, then left the 2 alone because you cant combine it with any other number.

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When Jennifer purchases her house, she pays a 3% realtor commission for the $1800,000 loan amount. How much commission did Jenni

ankoles [38]

Answer:

$54,000

Step-by-step explanation:

1,800,000 - 3% =

1,746,000

1,800,000 - 1,746,000 =

54,000

Jennifer paid $54,000 commission fee

8 0

1 year ago

Patricia is making two sizes of hamburgers: 0.5-pound andd 0.25-pound. How many of each size of hamburger could she make with 3.

mixas84 [53]

Answer:

She could make 5 of each type.

Step-by-step explanation:

If x is the number of 0.5 pound hamburgers and y is the number of 0.25 pound hamburgers:

0.5x + 0.25y = 3.75

Divide through by 0.25:

2x + y = 15

Some whole numbers which fit this relation are x = 5 and y = 5.

8 0

1 year ago

Read 2 more answers

PLEASE HELP!!!

labwork [276]

Answer:

it's B

Step-by-step explanation:

because I did this

7 0

2 years ago

Read 2 more answers

if each interior angle of a regular polygon measures 144 degrees, how many sides does the polygon have?

Lubov Fominskaja [6]

10 sides 180(10-2) 1800-360=1440/10 = 144

3 0

2 years 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

2 years ago