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RSB [31]
2 years ago
11

Given a1 = 8 and r = 2, identify the 23rd term

Mathematics
1 answer:
babymother [125]2 years ago
4 0

Answer: 33554432

Step-by-step explanation:

8 x (2 ^ (23 - 1)) = 33554432

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What is the formula for area of Parallelogram
Pachacha [2.7K]

Answer:

See below

Step-by-step explanation:

Area of parallelogram = base * height

5 0
3 years ago
Read 2 more answers
8-10 deal with the ambiguous SSA case. for each find all possible solutions and sketch the triangle in each case.
Artist 52 [7]
<span>For problem 8:

sin A/a sinB/b = sinC/c 

so sin50/15 = sinB/12 

0.766/15 =sinB/12 

0.051066 = sin B/12 

sin B = 0.6128 
B = 37.79 degrees 

sum of angles must equal 180 deg therefore C = 180-50-37.79 = 92.21 degrees and .766/15 = sin92.21/c 

.051066= 0.99925/c 
c = .99925/0.051066 = 19.567 

Problems 9 and 10 can also be done with the same method as problem 8.

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
</span>
3 0
3 years ago
Read 2 more answers
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
Please someone answer these 2 questions
saveliy_v [14]
1. C
2.A
Hope this helps, good luck :D
4 0
3 years ago
From an airplane flying over the ocean, the angle of depression to a submarine lying under the surface if 24 degrees 10 minutes.
forsale [732]
I think is 10 m/ s per sec
4 0
3 years ago
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