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Ber [7]
2 years ago
11

Find the 17th term of the of the arithmetic sequence-5,0,5,10

Mathematics
2 answers:
Lilit [14]2 years ago
8 0
An=a1+d(n-1)
first term is -5 (only way for it to make sense)
common difference is proabtly 5

an=-5+5(n-1)
17th term
an=-5+5(17-1)
an=-5+5(16)
an=-5+80
an=75
17th term is 75
kifflom [539]2 years ago
4 0
The n-th term for an arithmetic sequence with a common difference of 5 and a first term of -5 is given by
  a(n) = -5 + 5(n-1)
or
  a(n) = 5n -10

The 17th term is found by evaluating this for n=17.
  a17 = 5·17 - 10
  a17 = 75
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Ms. Li gave her students a quiz that was supposed to take 6 minutes to complete. She asked her students to record the amount of
Kamila [148]

B. I think it is b because i feel like a dot plot would be helpful

3 0
2 years ago
The hypotenuse of a right triangle is 50 and the short leg is 30, find the projection of the leg onto the hypotenuse
gladu [14]
Consider the attached figure. All the triangles shown are similar, so

CB/CA = CD/CB
30/50 = projection/30
projection = 30^2/50
projection = 18

6 0
2 years ago
Un globo vuela entre dos ciudades A y B, que distan entre sí 1.500 m. Los tripulantes del globo ven la ciudad A con un ángulo de
enot [183]

Answer:

La altura del globo con respecto al suelo es 449,6 metros.

Step-by-step explanation:

La afirmación está incompleta. El enunciado completo es: "Un globo vuela entre dos ciudades A y B, que distan entre sí 1.500 m. Los tripulantes del globo ven la ciudad A con un ángulo de depresión de 27°, mientras que para ver la ciudad B es de 36°. ¿Cuál es la altura aproximada del globo con respecto al suelo?

El diagrama geométrico de la situación se encuentra descrita en el archivo adjunto. La altura aproximada del globo puede obtenerse con ayuda de las funciones trigonométricas, en este caso, se recomienda utilizar la función tangente de los ángulos de depresión:

Ciudad A

\tan 27^{\circ} = \frac{h}{1500\,m-x}

0,510 = \frac{h}{1500\,m-x}

Ciudad B

\tan 36^{\circ} = \frac{h}{x}

0,727 = \frac{h}{x}

Donde h y x son la altura con respecto al suelo y la distancia horizontal con respecto a la ciudad A.

A continuación, se elimina la altura de ambas ecuaciones por igualación y se determina la distancia horizontal del globo con respecto a la ciudad A:

0,727\cdot x = 0,510\cdot (1500\,m-x)

1,237\cdot x = 765\,m

x = 618,432\,m

Finalmente, la altura del globo con respecto al suelo es:

h = 0,727\cdot x

h = 0,727\cdot (618,432\,m)

h = 449,600\,m

La altura del globo con respecto al suelo es 449,6 metros.

6 0
2 years ago
The ratio of the perimeter of square RSTU to the perimeter of square WXYZ is 1 to 3. The area of square RSTU is 36 square inches
Ainat [17]
√36 square inches = 6 inches (length of sq RSTU)
6 inches x 4=24 inches (per. of sq RSTU)
-----------------------------------
1u=24 inches
3u=24 inches x 3=72 inches (per. of sq WXYZ)
-----------------------------------
72 inches ÷ 4=18 inches (length of sq WXYZ)
18 inches x 18 inches = 324 square inches( area of sq WXYZ)

Ans: 324 square inches

5 0
2 years ago
Read 2 more answers
I've been stuck on this for so long and I have an exam soon, anybody who can help me :'( ?
san4es73 [151]
(i)  speed = distance / time
so time =  distance / speed
here we have

time t = 1080/x  hours

(ii) return flight  time  = 1080 / (x + 30)  hours

(a)  1080/x - 1080/(x + 30) = 1/2

Multiplying  through by the LCD 2x(x + 30) we get:-

1080*2(x + 30) - 2x*1080 = x(x+30)
2160x + 64800 - 2160x = x^2 + 30x
x^2 + 30x - 64800  = 0

(b)  factoring;  -64800 = 270 * -240  ans 270-240 = 30 so we have

(x + 270)(x - 240) = 0   so x = 240  ( we ignore the negative -270)

So the speed for outward journey is 240 km/hr

(c) time ffor outward flight = 1080 / 240 =  4 1/2  hours

(d) average speed for whole flight = distance / time
   Time for outward journey = 4.5 hours and time for  return journey = d / v
= 1080 / (240+30) =  4 hours
 Therefore the average speed for whole journey =  2160 / 8.5 = 254.1 km/hr
8 0
2 years ago
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