Hello !!!!!!Please help me ASAP

Find the missing lengths in the triangle, if y = 34.10.

joja [24]

Answer:

11.36

Step-by-step explanation:

3 0

3 years ago

What is the formula for area of a rectangle/square/parallelogram?

adell [148]

Answer:

Rectangle= lw (length times width)

Square= a²

Parallelogram=bh (base times height)

8 0

2 years ago

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

Find the squares of (y+3)

emmasim [6.3K]

Answer:

Since we know that

(a+b)

2

=a

2

+2ab+b

2

(a−b)

2

=a

2

−2ab+b

2

Now,

(i)

(x+3y)

2

=x

2

+2×x×3y+(3y)

2

=x

2

+6xy+9y

2

(ii)

(2x−5y)

2

=(2x)

2

−2×2x×5y+(5y)

2

=4x

2

−20xy+25y

2

(iii)

(a+

5a

1

)

2

=a

2

+2×a×

2a

1

+(

2a

1

)

2

=a

2

+1+

4a

2

1

(iv)

(2a−1a)=(1a)

2

(1a)

2

=1a

2

Answer

8 0

2 years ago

Help please:( i would appreciate it

Lady bird [3.3K]

The inverse would make the inputs the outputs (switching the x's and y's) because plugging in 4 into the inverse would somehow output 4 and 5, the inverse could not exist. plug in the inverse points to see that the vertical line test would fail.

8 0

3 years ago