Independent practice question 8=-12+5y

Each student in a class plays one of three sports soccer, volleyball or basketball. 1/4 of them play soccer. 3/5 of them play vo

Setler79 [48]

Answer: 3/20 of the class

Step-by-step explanation:

Add the amount of students that play soccer to amount of students that play volleyball

1/4 + 3/5 = 17/20

Subtract 17/20 (soccer and volley ball) from 20/20 or 1

20/20 - 17/20 = 3/20

7 0

3 years ago

Which set of measurements could represent the three sides of a triangle?

yawa3891 [41]

Answer:

The side lengths of a right triangle is 11cm, 60cm and 61cm, that could be selected from the given measurements.

Step-by-step explanation:

The measurements are,

7cm, 11cm, 54cm, 60cm, 61cm, 65cm

Step:1

To check the right angle triangle, Pythagorean theorem can be used.

For a Pythagorean theorem,

..........................(1)

The side values are lower than the hypotenuse,

...................................(2)

Where,

a,b - side values

c - Hypotenuse

For right angle triangle, c > a, b

Alternative : 1

Take, a = 7cm, b = 11cm

From eqn (2),

= = 13.04

The above value is not equal to the any one of the values of ( 54cm. 60cm, 61cm, 65cm ), So its not an sides of right triangle.

Alternative : 2

Take, a = 7cm, b = 54cm

From eqn (2),

= = 54.45

The above value is not equal to the any one of the values of (60cm, 61cm, 65cm ), So its not an sides of right triangle.

Alternative : 3

Take, a = 7cm, b = 60cm

From eqn (2),

= = 60.406

The above value is not equal to the any one of the values of (61cm, 65cm ), So its not an sides of right triangle.

Alternative : 4

Take, a = 7cm, b = 61cm

From eqn (2),

= = 61.40

The above value is not equal to the values of (65cm ), So its not an sides of right triangle.

Alternative : 5

Take, a = 11cm, b = 54cm

From eqn (2),

= = 55.1089

The above value is not equal to the any one of the values of (60cm, 61cm, 65cm ), So its not an sides of right triangle.

Alternative : 6

Take, a = 11cm, b = 60cm

From eqn (2),

= = 61

The above value is equal to the values of (61cm ), So its an sides of right triangle. The three sides are 11, 60 and 61.

Step:2

Check for solution,

Result:

The side lengths of a right triangle is 11cm, 60cm and 61cm, that could be selected from the given measurements.

Step-by-step explanation: The side lengths of a right triangle is 11cm, 60cm and 61cm.

4 0

3 years ago

I need help can u help me please and thanks

son4ous [18]

Answer:

17.74

Step-by-step explanation:

The fraction is easily converted to one with a denominator of 100 (a power of 10):

$\dfrac{37}{50}=\dfrac{37\cdot 2}{50\cdot 2}=\dfrac{74}{100}=0.74$

Then the number of interest is ...

17 37/50 = 17 74/100 = 17.74

3 0

3 years ago

Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by the given curves

Vadim26 [7]

The expression on the left side describes a parabola. Factorize it to determine where it crosses the y-axis (i.e. the line x = 0) :

-3y² + 9y - 6 = -3 (y² - 3y + 2)

… = -3 (y - 1) (y - 2) = 0

⇒ y = 1 or y = 2

Also, complete the square to determine the vertex of the parabola:

-3y² + 9y - 6 = -3 (y² - 3y) - 6

… = -3 (y² - 3y + 9/4 - 9/4) - 6

… = -3 (y² - 2•3/2 y + (3/2)²) + 27/4 - 6

… = -3 (y - 3/2)² + 3/4

⇒ vertex at (x, y) = (3/4, 3/2)

I've attached a sketch of the curve along with one of the shells that make up the solid. For some value of x in the interval 0 ≤ x ≤ 3/4, each cylindrical shell has

radius = x

height = y⁺ - y⁻

where y⁺ refers to the half of the parabola above the line y = 3/2, and y⁻ is the lower half. These halves are functions of x that we obtain from its equation by solving for y :

x = -3y² + 9y - 6

x = -3 (y - 3/2)² + 3/4

x - 3/4 = -3 (y - 3/2)²

-x/3 + 1/4 = (y - 3/2)²

± √(1/4 - x/3) = y - 3/2

y = 3/2 ± √(1/4 - x/3)

y⁺ and y⁻ are the solutions with the positive and negative square roots, respectively, so each shell has height

(3/2 + √(1/4 - x/3)) - (3/2 - √(1/4 - x/3)) = 2 √(1/4 - x/3)

Now set up the integral and compute the volume.

$\displaystyle 2\pi \int_{x=0}^{x=3/4} 2x \sqrt{\frac14 - \frac x3} \, dx$

Substitute u = 1/4 - x/3, so x = 3/4 - 3u and dx = -3 du.

$\displaystyle 2\pi \int_{u=1/4-0/3}^{u=1/4-(3/4)/3} 2\left(\frac34 - 3u\right) \sqrt{u} \left(-3 \, du\right)$

$\displaystyle -12\pi \int_{u=1/4}^{u=0} \left(\frac34 - 3u\right) \sqrt{u} \, du$

$\displaystyle 12\pi \int_{u=0}^{u=1/4} \left(\frac34 u^{1/2} - 3u^{3/2}\right) \, du$

$\displaystyle 12\pi \left(\frac34\cdot\frac23 u^{3/2} - 3\cdot\frac25u^{5/2}\right) \bigg|_{u=0}^{u=1/4}$

$\displaystyle 12\pi \left(\frac12 u^{3/2} - \frac65u^{5/2}\right) \bigg|_{u=0}^{u=1/4}$

$\displaystyle 12\pi \left(\frac12 \left(\frac14\right)^{3/2} - \frac65\left(\frac14\right)^{5/2}\right) - 12\pi (0 - 0)$

$\displaystyle 12\pi \left(\frac1{16} - \frac3{80}\right) = \frac{12\pi}{40} = \boxed{\frac{3\pi}{10}}$

6 0

2 years ago

How do I find the area of the base of a three dimensional cone

notka56 [123]

Idkk dmh shi fw food demic

8 0

4 years ago