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Stolb23 [73]
2 years ago
12

What is the area of this figure?

Mathematics
1 answer:
Luden [163]2 years ago
3 0

Answer:

The area of this figure is 24mm.

Step-by-step explanation:

4mm*3mm=12mm

1mm*6mm=6mm

3mm*4mm=12mm/2mm=6mm

6mm+6mm+12mm=24mm

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A $250 gaming system is discounted by 25%. It was then discounted another 10% because the box was opened by a customer. a. The s
iris [78.8K]

Answer:

a. Sale Price=\$168.75

b. Total percent discount=32.5\%

Step-by-step explanation:

First Discount:

Initial Price=\$250

Discount=25\%

25\%\ of\ 250=\frac{25}{100}\times 250=62.5\\\\Now\ Price=250-62.5=$187.5

Second Discount:

Price=187.5\\\\Discount=10\%\\\\10\%\ of\ 187.5=\frac{10}{100}\times 187.5=18.75\\\\Now\ Sale\ Price=187.5-18.75=\$168.75

Effective Discount:

Initial Price=\$250

Final Price=\$168.75

Total discount=250-168.75=81.25

\%\ discount=\frac{81.25}{250}\times 100=32.5\%

3 0
3 years ago
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
En la tienda de mascotas "Animalo-T", se desea elevar un elefante de 2,900 kg utilizando una elevadora hidráulica de plato grand
Korvikt [17]

Answer:

Se requiere una fuerza de 854.473 newtons sobre el émbolo pequeño.

Step-by-step explanation:

Por el Principio de Pascal se conoce que el esfuerzo experimentado por el elefante es igual a la presión ejercida por el plato pequeño. Es decir:

\frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}} (1)

Donde:

F_{1} - Fuerza experimentada por el elefante, medida en newtons.

F_{2} - Fuerza aplicada sobre el plato pequeño, medida en newtons.

A_{1} - Área del plato grande, medida en metros cuadrados.

A_{2} - Área del plato pequeño, medida en metros cuadrados.

La fuerza aplicada sobre el plato pequeño es:

F_{2} = \left(\frac{A_{2}}{A_{1}} \right)\cdot F_{1}

La fuerza experimentada por el elefante es su propio peso. Por otra parte, el área del plato es directamente proporcional al cuadrado de su diámetro. Es decir:

F_{2} = \left(\frac{D_{2}}{D_{1}} \right)^{2}\cdot m\cdot g (2)

Donde:

D_{1} - Diámetro del plato grande, medido en centímetros.

D_{2} - Diámetro del plato pequeño, medido en centímetros.

m - Masa del elefante, medida en kilogramos.

g - Aceleración gravitacional, medida en metros por segundo cuadrado.

Si sabemos que D_{1} = 0.75\,m, D_{2} = 0.13\,m, m = 2900\,kg y g = 9.807\,\frac{m}{s^{2}}, entonces la fuerza a aplicar al émbolo pequeño es:

F_{2} = \left(\frac{0.13\,m}{0.75\,m} \right)^{2}\cdot (2900\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)

F_{2} = 854.473\,N

Se requiere una fuerza de 854.473 newtons sobre el émbolo pequeño.

3 0
3 years ago
I need help on these problems
Mazyrski [523]
Hope this helps and hope you can read it, if not let me know :)

4 0
3 years ago
Identify the type of function represented by
SCORPION-xisa [38]
That's definitely an example of exponential decay, since the base (1/2) (also called the "common ratio") is greater than 0 but less than 1.
6 0
3 years ago
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