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3 years ago

The first two terms of an arithmetic sequence are a1 = 2 and a2 = 4. What is a10,the tenth term?

2 answers:

7 0

A(1) = 2
a(2) = 4

You can see that the terms are increasing by +2.
so

a(1) = 2
a(2) = 4
a(3) = 6
a(4) = 8
a(5) = 10
a(6) = 12
a(7) = 14
a(8) = 16
a(9) = 18
a(10) = 20

your answer is a(10) = 20.

rosijanka [135]3 years ago

5 0

Answer:

Step-by-step explanation:

a1 = 2

a2 = 4

a3 = 6

........

a10 = 20

Multiply the subscript by 2 and you'll get your answer.

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Please please help me

Gekata [30.6K]

Answer:

$\large\boxed{\dfrac{\boxed{8}}{81}}$

Step-by-step explanation:

$\text{If}\ a_1,\ a_2,\ a_3,\ a_4,\ ...,\ a_n\ \text{is the geometric sequence, then}\\\\\dfrac{a_2}{a_1}=\dfrac{a_3}{a_2}=\dfrac{a_4}{a_3}=...=\dfrac{a_n}{a_{n-1}}=constant=r-\text{common ration}.\\\\\text{We have}\ \dfrac{1}{2},\ \dfrac{1}{3},\ \dfrac{2}{9},\ \dfrac{4}{27},\ ...\\\\\dfrac{\frac{1}{3}}{\frac{1}{2}}=\dfrac{1}{3}\cdot\dfrac{2}{1}=\dfrac{2}{3}\\\\\dfrac{\frac{2}{9}}{\frac{1}{3}}=\dfrac{2}{9}\cdot\dfrac{3}{1}=\dfrac{2}{3}\\\\\dfrac{\frac{4}{27}}{\frac{2}{9}}=\dfrac{4}{27}\cdot\dfrac{9}{2}=\dfrac{2}{3}$

$\bold{CORRECT}\\\\\dfrac{x}{\frac{4}{27}}=\dfrac{2}{3}\qquad\text{multiply both sides by}\ \dfrac{4}{27}\\\\x=\dfrac{2}{3}\cdot\dfrac{4}{27}\\\\x=\dfrac{8}{81}$

8 0

3 years ago

Jane and sali cycled along the same 63 km route.

Alina [70]

Answer:

<em>Sali's speed was 18.75 km/h.</em>

Step-by-step explanation:

Jane took 3.5 hours to cycle the 63 km.

As, $Speed= \frac{Distance}{Time}$ , so the speed of Jane will be: $\frac{63}{3.5} km/h = 18 km/h$

Suppose, the speed of Sali is $x$ km/h

Sali caught up with Jane when they had both cycled 30 km.

So, <u>the time required for Jane to cycle 30 km</u> $= \frac{30}{18}=\frac{5}{3} hours$ and <u>the time required for Sali to cycle 30 km</u> $=\frac{30}{x} hours$

Given that, Sali started to cycle 4 minutes or $(\frac{4}{60}) or (\frac{1}{15}) hours$ after Jane started to cycle. So, the equation will be.......

$\frac{5}{3}-\frac{30}{x}= \frac{1}{15}\\ \\ \frac{30}{x}= \frac{5}{3}-\frac{1}{15}\\ \\ \frac{30}{x}= \frac{24}{15}\\ \\ 24x=450\\ \\ x= \frac{450}{24}= 18.75$

Thus, the speed of Sali was 18.75 km/h.

8 0

3 years ago

Solve the system of equations - 8x + 3y = -17 and 3x - y = 7 by combining the<br> equlations.

NeTakaya

Given :

Two equations :

-8x + 3y = -17 ....1)

3x - y = 7 ....2)

To Find :

The solution of the system.

Solution :

Multiplying equation 2) by 3 and adding with equation 1), we get :

(-8x + 3y) + 3(3x - y) = -17 + 21

x = 4

Putting above value of x equation 2) we get :

12 - y = 7

y = 5

Therefore, solution of system is ( 4,5 ).

8 0

3 years ago

5/8+3/4 / -2/3 - 5/6 =

m_a_m_a [10]

Answer:

-11/12

Step-by-step explanation:

Add 5/8+3/4 (6/8) which is 11/8 and then get -2/3 - 5/6 which simplifies to -3/2

Multiply -3/2 by 4 to get it to like terms

11/8 / -12/8

Multiply and then simplify.

If you need help, just comment!!

3 0

3 years ago

At an airport, 76% of recent flights have arrived on time. A sample of 11 flights is studied. Find the probability that no more

I am Lyosha [343]

Answer:

The probability is $P( X \le 4 ) = 0.0054$

Step-by-step explanation:

From the question we are told that

The percentage that are on time is p = 0.76

The sample size is n = 11

Generally the percentage that are not on time is

$q = 1- p$

$q = 1- 0.76$

$q = 0.24$

The probability that no more than 4 of them were on time is mathematically represented as

$P( X \le 4 ) = P(1 ) + P(2) + P(3) + P(4)$

=> $P( X \le 4 ) = \left n } \atop {}} \right.C_1 p^{1} q^{n- 1} + \left n } \atop {}} \right.C_2p^{2} q^{n- 2} + \left n } \atop {}} \right.C_3 p^{3} q^{n- 3} + \left n } \atop {}} \right.C_4 p^{4} q^{n- 4}$

$P( X \le 4 ) = \left 11 } \atop {}} \right.C_1 p^{1} q^{11- 1} + \left 11 } \atop {}} \right.C_2p^{2} q^{11- 2} + \left 11 } \atop {}} \right.C_3 p^{3} q^{11- 3} + \left 11 } \atop {}} \right.C_4 p^{4} q^{11- 4}$

$P( X \le 4 ) = \left 11 } \atop {}} \right.C_1 p^{1} q^{10} + \left 11 } \atop {}} \right.C_2p^{2} q^{9} + \left 11 } \atop {}} \right.C_3 p^{3} q^{8} + \left 11 } \atop {}} \right.C_4 p^{4} q^{7}$

$= \frac{11! }{ 10! 1!} (0.76)^{1} (0.24)^{10} + \frac{11!}{9! 2!} (0.76)^2 (0.24)^{9} + \frac{11!}{8! 3!} (0.76)^{3} (0.24)^{8} + \frac{11!}{7!4!} (0.76)^{4} (0.24)^{7}$

$P( X \le 4 ) = 0.0054$

4 0

3 years ago