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Black_prince [1.1K]
2 years ago
7

Find the area of the following ellipse (round to nearest tenth).

Mathematics
1 answer:
goblinko [34]2 years ago
8 0

Answer:

\huge\boxed{A=60\pi cm^2}

Step-by-step explanation:

The formula of an area of a elipse:

A=\pi ab

We have:

2a=10cm\to a=5cm\\\\2b=24cm\to b=12cm

Substitute:

A=\pi\dot5\cdot12=60\pi(cm^2)

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WITCHER [35]

Answer:

$110

Step-by-step explanation:

1. The total cost:

<em>(3 * 65) + (2 * 70) + (1 * 55)</em> -> cost of dresses + jeans + earrings

Total: 195 + 140 + 55 = <em>390</em>.

2. Money leftover:

<em>500 - 390</em> = $110.

4 0
3 years ago
What is the volume of the cylinder if the height is 2x+7 and the radius is x-3
shusha [124]

Answer:

Therefore the volume of the cylinder=\pi(2x^3-5x^2-24x+63)  cubic units

Step-by-step explanation:

Given height of a cylinder is (h) =2x+7

and radius  (r)= x-3

Volume of the cylinder = \pi r^2 h

                                  =\pi (x-3)^2(2x+7)

                                  =\pi (x^2-6x+9)(2x+7)

                                  =\pi (2x^3-12x^2+18x+7x^2-42x+63)

                                  =\pi(2x^3-5x^2-24x+63)  cubic units

Therefore the volume of the cylinder=\pi(2x^3-5x^2-24x+63)  cubic units

5 0
3 years ago
Help help math suckssss
stiv31 [10]

Answer:

12, 20

Step-by-step explanation:

3/8 of 32 is 12. She ate 3/8, so she ate 12 apples. The question wants to know how many are left over, so you subtract 12 from 32 and get 20.

5 0
3 years ago
Una lancha que viaja a 10 m/s pasa por debajo de un puente 3 segundos después que ha pasado un bote que viaja a 7 m/s, ¿después
ExtremeBDS [4]

Answer:

La lancha y el bote se encontrarán a 70 metros de distancia del puente.

Step-by-step explanation:

Sea el punto debajo del puente el punto de referencia y que ambas lanchas se desplazan a velocidad a continuación, las ecuaciones cinemáticas para cada embarcación son presentadas a continuación:

Bote a 7 metros por segundo

x_{A} = x_{o}+v_{A}\cdot t (Ec. 1)

Lancha a 10 metros por segundo

x_{B} = x_{o}+v_{B}\cdot (t-3\,s) (Ec. 2)

Donde:

x_{o} - Posición debajo del puente, medido en metros.

x_{A}, x_{B} - Posición final de cada embarcación, medido en metros.

v_{A}, v_{B} - Velocidad de cada embarcación, medida en metros por segundo.

t - Tiempo, medido en segundos.

Para determinar la posición en la que ambas embarcaciones se encuentran, se debe determinar el instante en que ocurre a partir de la siguiente condición: x_{A} = x_{B}

Igualando (Ec. 1) y (Ec. 2) se tiene que:

v_{A}\cdot t = v_{B}\cdot (t-3\,s)

Ahora despejamos el tiempo:

3\cdot v_{B} = (v_{B}-v_{A})\cdot t

t = \frac{3\cdot v_{B}}{v_{B}-v_{A}}

Si sabemos que v_{B} = 10\,\frac{m}{s} y v_{A} = 7\,\frac{m}{s}, entonces:

t = \frac{3\cdot \left(10\,\frac{m}{s} \right)}{10\,\frac{m}{s}-7\,\frac{m}{s}}

t = 10\,s

Ahora, la posición de encuentro es: (x_{o} = 0\,m, v_{A} = 7\,\frac{m}{s} y t = 10\,s)

x_{A} = 0\,m + \left(7\,\frac{m}{s} \right)\cdot (10\,s)

x_{A} = 70\,m

La lancha y el bote se encontrarán a 70 metros de distancia del puente.

6 0
3 years ago
Someonee can u tell ne​
Vesnalui [34]

Answer:

35

Step-by-step explanation:

i did the sqrt of 1225 and got 35 so there should be 35 rows each with 25 plants

4 0
3 years ago
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