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natka813 [3]
2 years ago
7

Point B is at (3,1).

Mathematics
1 answer:
Alik [6]2 years ago
6 0

Answer:

A(1,−1) and C(5,3)

Step-by-step explanation:

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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
Grade 12<br>mathematics<br>(√98-√50)² ​
allsm [11]

Answer:

8

Step-by-step explanation:

Two different approaches:

<u>Method 1</u>

Apply radical rule √(ab) = √a√b to simplify the radicals:

√98 = √(49 x 2) = √49√2 = 7√2

√50 = √(25 x 2) = √25√2 = 5√2

Therefore,  (√98 - √50)² = (7√2 - 5√2)²

                                         = (2√2)²

                                         = 4 x 2

                                         = 8

<u>Method 2</u>

Use the perfect square formula: (a - b)² = a² - 2ab + b²

where a = √98   and   b = √50

So            (√98 - √50)² = (√98)² - 2√98√50 + (√50)²

                                      = 98 - 2√98√50 + 50

                                      = 148 - 2√98√50

Apply radical rule √(ab) = √a√b to simplify radicals:

√98 = √(49 x 2) = √49√2 = 7√2

√50 = √(25 x 2) = √25√2 = 5√2

Therefore,     148 - 2√98√50 = 148 - (2 × 7√2 × 5√2)

                                                 = 148 - 140

                                                 = 8

4 0
2 years ago
The perimeter of a school gym is 522 feet. The gym measures 80 feet wide. Determine the length of the school. PLEASE EXPLAIN THE
lidiya [134]
Total perimeter is 522 ft. I will guess the gym is a big rectangle. 

Therefore, Perimeter of a rectangle = 2L + 2W
P= 2L+ 2W

We know 80 ft wide. so now we plug and chug the equation we came up with

P=522 ft
P=2L+ 2(80) 

522=2L+ 160

L= 181

hope that helps

3 0
3 years ago
(2+5x)9= combine like terms
rodikova [14]

Answer:

18+45x

Step-by-step explanation:

4 0
3 years ago
Find the lateral surface area of the cylinder. Round your answer to the nearest tenth.
zheka24 [161]

Answer:

The answer is "The second choice".

Step-by-step explanation:

r=6\\\\h=13 \\\\

Formula:

A=2\pi rh

=2\times 3.14 \times 6 \times 13\\\\=2\times 3.14 \times 78\\\\=3.14 \times 156\\\\=3.14 \times 156\\\\=489.84 \approx 489.8 ft^2

5 0
2 years ago
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