All categories

3 years ago

PLZZZZZZZZZZZZ PLZZZZZZZZZZZZ PLZZZZZZZZZZZZ PLZZZZZZZZZZZZ help

Mathematics

2 answers:

Tpy6a [65]3 years ago

4 0

A. Decreased by 15% compared with last years amount

pogonyaev3 years ago

3 0

Answer:

a. option is the correct answer

You might be interested in

Find the area of a circle with radius 6 m.

Vera_Pavlovna [14]

The right answer is 113.04m^2

please see the attached picture for full solution

Hope it helps

3 0

3 years ago

Pleas help me on this

labwork [276]

Answer:

Step-by-step explanation:

x+x+x

=3x

if i am wrong than sorry

8 0

2 years ago

Read 2 more answers

Use cylindrical coordinates to evaluate the triple integral ∭ where E is the solid bounded by the circular paraboloid z = 9 - 16

4vir4ik [10]

Answer:

$\mathbf{\iiint_E E \sqrt{x^2+y^2} \ dV =\dfrac{81 \ \pi}{80}}$

Step-by-step explanation:

The Cylindrical coordinates are:

x = rcosθ, y = rsinθ and z = z

From the question, on the xy-plane;

$9 -16 (x^2 + y^2) = 0 \\ \\ 16 (x^2 + y^2) = 9 \\ \\ x^2+y^2 = \dfrac{9}{16}$

$x^2+y^2 = (\dfrac{3}{4})^2$

where:

0 ≤ r ≤ $\dfrac{3}{4}$ and 0 ≤ θ ≤ 2π

∴

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} \int ^{9-16r^2}_{0} \ r \times rdzdrd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 z|^{z= 9-16r^2}_{z=0} \ \ \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 ( 9-16r^2}) \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} ( 9r^2-16r^4}) \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( \dfrac{9r^3}{3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0} \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( 3r^3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0} \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) d \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) \theta |^{2 \pi}_{0}$

$\iiint_E E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}})2 \pi$

$\iiint_E E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{64}}-\dfrac{243}{320}})2 \pi$

$\iiint_E E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{160}})2 \pi$

$\mathbf{\iiint_E E \sqrt{x^2+y^2} \ dV =\dfrac{81 \ \pi}{80}}$

4 0

3 years ago

Consider the numbers 4/15 and 3/10 (a) Compare the numbers. (B) find a fraction between the two given numbers

valentinak56 [21]

As a decimal 4/15 = 0.2666667
As a decimal 3/10 = 0.3 [talk about close]
3/10 is bigger.

You could take an average to find something between 4/15 and 3/10
1/2(4/15 + 2/10)
1/2(8/30 + 6/30)
1/2(14/30)
1/2(7/15)
7/30 is between the two given numbers..

5 0

3 years ago

On a scale drawing , a tree is 6 3/4 inches tall.The scale factor is 1/20 . Find the height of the actual tree.

Crank

Answer:

Step-by-step explanation: ⣠⠤⠖⠚⢉⣩⣭⡭⠛⠓⠲⠦⣄⡀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⢀⡴⠋⠁⠀⠀⠊⠀⠀⠀⠀⠀⠀⠀⠀⠀⠉⠳⢦⡀⠀⠀⠀⠀

⠀⠀⠀⠀⢀⡴⠃⢀⡴⢳⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠙⣆⠀⠀⠀

⠀⠀⠀⠀⡾⠁⣠⠋⠀⠈⢧⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠈⢧⠀⠀

⠀⠀⠀⣸⠁⢰⠃⠀⠀⠀⠈⢣⡀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠈⣇⠀

⠀⠀⠀⡇⠀⡾⡀⠀⠀⠀⠀⣀⣹⣆⡀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢹⠀

⠀⠀⢸⠃⢀⣇⡈⠀⠀⠀⠀⠀⠀⢀⡑⢄⡀⢀⡀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢸⡇

⠀⠀⢸⠀⢻⡟⡻⢶⡆⠀⠀⠀⠀⡼⠟⡳⢿⣦⡑⢄⠀⠀⠀⠀⠀⠀⠀⠀⢸⡇

⠀⠀⣸⠀⢸⠃⡇⢀⠇⠀⠀⠀⠀⠀⡼⠀⠀⠈⣿⡗⠂⠀⠀⠀⠀⠀⠀⠀⢸⠁

⠀⠀⡏⠀⣼⠀⢳⠊⠀⠀⠀⠀⠀⠀⠱⣀⣀⠔⣸⠁⠀⠀⠀⠀⠀⠀⠀⢠⡟⠀

⠀⠀⡇⢀⡇⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠠⠀⡇⠀⠀⠀⠀⠀⠀⠀⠀⢸⠃⠀

⠀⢸⠃⠘⡇⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢸⠁⠀⠀⢀⠀⠀⠀⠀⠀⣾⠀⠀

⠀⣸⠀⠀⠹⡄⠀⠀⠈⠁⠀⠀⠀⠀⠀⠀⠀⡞⠀⠀⠀⠸⠀⠀⠀⠀⠀⡇⠀⠀

⠀⡏⠀⠀⠀⠙⣆⠀⠀⠀⠀⠀⠀⠀⢀⣠⢶⡇⠀⠀⢰⡀⠀⠀⠀⠀⠀⡇⠀⠀

⢰⠇⡄⠀⠀⠀⡿⢣⣀⣀⣀⡤⠴⡞⠉⠀⢸⠀⠀⠀⣿⡇⠀⠀⠀⠀⠀⣧⠀⠀

⣸⠀⡇⠀⠀⠀⠀⠀⠀⠉⠀⠀⠀⢹⠀⠀⢸⠀⠀⢀⣿⠇⠀⠀⠀⠁⠀⢸⠀⠀

⣿⠀⡇⠀⠀⠀⠀⠀⢀⡤⠤⠶⠶⠾⠤⠄⢸⠀⡀⠸⣿⣀⠀⠀⠀⠀⠀⠈⣇⠀

⡇⠀⡇⠀⠀⡀⠀⡴⠋⠀⠀⠀⠀⠀⠀⠀⠸⡌⣵⡀⢳⡇⠀⠀⠀⠀⠀⠀⢹⡀

⡇⠀⠇⠀⠀⡇⡸⠁⠀⠀⠀⠀⠀⠀⠀⠀⠀⠙⠮⢧⣀⣻⢂⠀⠀⠀⠀⠀⠀⢧

⣇⠀⢠⠀⠀⢳⠇⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠈⡎⣆⠀⠀⠀⠀⠀⠘

⢻⠀⠈⠰⠀⢸⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠰⠘⢮⣧⡀⠀⠀⠀⠀

⠸⡆⠀⠀⠇⣾⠀⠀⠀⠀⠀⠀⠀⠀⠀⢠⠆⠀⠀⠀⠀⠀⠀⠀⠙⠳⣄⡀⢢⡀

8 0

2 years ago