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2 years ago

In the given figure, values of x and y, are 1. 40°, 25° 2. 30°, 40° 3. 30°, 20° 4. 40°, 35°

Mathematics

1 answer:

Virty [35]2 years ago

6 0

Answer:

1. 40°, 25°

Step-by-step explanation:

The sum of opposite angles of a quadrilateral inscribed in a circle is 180°.

2x + 4 + 4y - 4 = 180

x+ 10 + 5y + 5 = 180

2x + 4y = 180

x +5y + 15 = 180

x + 2y = 90

x + 5y = 165

x = 90 - 2y

x = 165 - 5y

90 - 2y = 165 - 5y

5y - 2y = 165 - 90

3y = 75

y = 25

x = 90- 2y = 90 - 2*25 = 40

x= 40, y = 25

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a string was 4m long. she gave 1/4m of it to her sister and 3/4m to her friends. how much string had she given away?

OverLord2011 [107]

The correct answer is: " 1 m " .
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Explanation:
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Add up the amounts she had given away:

3/4 m + 1/4 m = 4/4 m = 1 m.
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The answer is: " 1 m " .
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6 0

3 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

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3 years ago

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