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2 years ago

Write the equation of a line that is perpendicular to y= -1 and passes through the point (8,-4)

Mathematics

1 answer:

satela [25.4K]2 years ago

3 0

Answer:

y=-1-12

I hope you get it right

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A cone with base area 16 pi square and height 9 feet

alekssr [168]

Area of cone = 1/3 base area x height
Area of cone = 1/3 (16π) x 9
Area of cone = 48π ft³ or 150.8 ft³

8 0

2 years ago

Find z for the following proportion:<br><br><br> z/9= 2/3

serg [7]

Cross multiplication will help find the value of x
$\frac{z}{9} = \frac{2}{3} 


$
9*2=18
3*z=3z

Now we have :
3z=18
Divide both sides by 3 to isolate z

Final answer: z=6

4 0

2 years ago

What is the value of 3^3 - (4^2 - 2^3)?

Illusion [34]

3³ - (4² - 2³)

3³ = 3x3x3 = 27

4² = 4x4 = 16

2³ = 2x2x2 = 8

3³ - (4² - 2³) = 27 - (16 - 8) = 27 - 16 + 8 = 19 (answer C)

4 0

2 years ago

Read 2 more answers

5x + 8 - 7x = -4x + 1

irakobra [83]

Step-by-step explanation:

5x+8-7x=4x+1

5x+7x-4x=8+1

8x=9

6 0

2 years ago

Read 2 more answers

Find the mass of the solid paraboloid Dequals={(r,thetaθ,z): 0less than or equals≤zless than or equals≤8181minus−r2, 0less th

Lubov Fominskaja [6]

Answer:

M = 5742π

Step-by-step explanation:

Given:-

- Find the mass of a solid with the density ( ρ ):

ρ ( r, θ , z ) = 1 + z / 81

- The solid is bounded by the planes:

0 ≤ z ≤ 81 - r^2

0 ≤ r ≤ 9

Find:-

Find the mass of the solid paraboloid

Solution:-

- The mass (M) of any solid body is given by the following triple integral formulation:

$M = \int \int \int {p ( r ,theta, z)} \, dV\\\\$

- We can write the above expression in cylindrical coordinates:

$M = \int\limits\int\limits_r\int\limits_z {r*p(r,theta,z)} \, dz.dr.dtheta \\\\M = \int\limits\int\limits_r\int\limits_z {r*[ 1 + \frac{z}{81}] } \, dz.dr.dtheta\\\\$

- Perform integration:

$M = \int\limits\int\limits_r{r*[ z + \frac{z^2}{162}] } \,|_0^8^1^-^r^2 dr.dtheta\\\\M = \int\limits\int\limits_r{r*[ 81-r^2 + \frac{(81-r^2)^2}{162}] } \, dr.dtheta\\\\M = \int\limits\int\limits_r{r*[ 81-r^2 + \frac{6561 -162r + r^2}{162}] } \, dr.dtheta\\\\M = \int\limits\int\limits_r{r*[ 81-r^2 + 40.5 -r +\frac{r^2}{162} ] } \, dr.dtheta\\\\M = \int\limits\int\limits_r{[ 121.5r-r^2 -\frac{161r^3}{162} ] } \, dr.dtheta\\\\$

$M = 2*\int\limits_0^\pi {[ 121.5r^2-r^3 -\frac{161r^4}{162} ] } |_0^6 \, dtheta\\\\M = 2*\int\limits_0^\pi {[ 121.5(6)^2-(6)^3 -\frac{161(6)^4}{162} ] } \, dtheta\\\\M = 2*\int\limits_0^\pi {[ 4375-216 -1288] } \, dtheta\\\\M = 2*\int\limits_0^\pi {[ 2871] } \, dtheta\\\\M = 5742\pi kg$

- The mass evaluated is M = 5742π

8 0

2 years ago