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3 years ago

Calculate the next 3 terms and write the formula for the nth term for the following sequences. x, x+2, x+4,…

Mathematics

2 answers:

Vikki [24]3 years ago

8 0

Answer:

Forth term = x + 6

Fifth term = x + 8

Sixth term = x + 10

Formular for nth term:

${ \tt{n _{th} = a + (n - 1)d}} \\ { \tt{{n _{th} = x + (n - 1) \times 2}}} \\ { \tt{{n _{th} =x + 2n - 2 }}}$

sergey [27]3 years ago

5 0

Answer:

Hello,

Step-by-step explanation:

First term of this aritmetical sequence is x+0*2

$U_1=x+0*2\\U_2=x+2=x+1*2\\U_3=x+4=x+2*2\\U_4=x+6=x+3*2\\\\n^{th}\ term\ is\ U_n=x+(n-1)*2\\$

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Which angles are neither obtuse angles nor acute angles?

sleet_krkn [62]

Answer:

A right angle is neither Obtuse nor Acute

Step-by-step explanation:

5 0

2 years ago

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Kristie has taken five tests in science class. The average of all five of Kristie's test scores is 94. The average of her last t

Bingel [31]

Answer: The average of the first two test scores is 97.

Step-by-step explanation: Given that Kristie has taken five tests in science class. The average of all five of Kristie's test scores is 94 and the average of her last three test scores is 92.

We are to find the average score of her first two tests.

Let a1, a2, a3, a4 and a5 be teh scores of Kristle in first , second, third, fourth and fifth tests respectively.

Then, according to the given information, we have

$\dfrac{a1+a2+a3+a4+a5}{5}=94\\\\\Rightarrow a1+a2+a3+a4+a5=94\times5\\\\\Rightarrow a1+a2+a3+a4+a5=470~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

and

$\dfrac{a3+a4+a5}{3}=92\\\\\Rightarrow a3+a4+a5=92\times3\\\\\Rightarrow a3+a4+a5=276~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

Subtracting equation (ii) from equation (i), we get

$(a1+a2+a3+a4+a5)-(a3+a4+a5)=470-276\\\\\Rightarrow a1+a2=194\\\\\Rightarrow \dfrac{a1+a2}{2}=\dfrac{194}{2}\\\\\Rightarrow \dfrac{a1+a2}{2}=97.$

Thus, the average of the first two test scores is 97.

6 0

3 years ago

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains liters of a dye solution with a

Alja [10]

Answer:

t = 460.52 min

Step-by-step explanation:

Here is the complete question

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 liters of a dye solution with a concentration of 1 g/liter. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 liters/min, the well-stirred solution flowing out at the same rate.Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.

Solution

Let Q(t) represent the amount of dye at any time t. Q' represent the net rate of change of amount of dye in the tank. Q' = inflow - outflow.

inflow = 0 (since the incoming water contains no dye)

outflow = concentration × rate of water inflow

Concentration = Quantity/volume = Q/200

outflow = concentration × rate of water inflow = Q/200 g/liter × 2 liters/min = Q/100 g/min.

So, Q' = inflow - outflow = 0 - Q/100

Q' = -Q/100 This is our differential equation. We solve it as follows

Q'/Q = -1/100

∫Q'/Q = ∫-1/100

㏑Q = -t/100 + c

$Q(t) = e^{(-t/100 + c)} = e^{(-t/100)}e^{c} = Ae^{(-t/100)}\\Q(t) = Ae^{(-t/100)}$

when t = 0, Q = 200 L × 1 g/L = 200 g

$Q(0) = 200 = Ae^{(-0/100)} = Ae^{(0)} = A\\A = 200.\\So, Q(t) = 200e^{(-t/100)}$

We are to find t when Q = 1% of its original value. 1% of 200 g = 0.01 × 200 = 2

$2 = 200e^{(-t/100)}\\\frac{2}{200} = e^{(-t/100)}$

㏑0.01 = -t/100

t = -100㏑0.01

t = 460.52 min

6 0

3 years ago

Help please!

Lemur [1.5K]

Answer:

Step-by-step explanation:

a I’m pretty sure

5 0

3 years ago

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21. what are the excluded values of the function? y= 5/6x-72

tatiyna

you have to calculate the SID number and then take it upon yourself and divided X it's actual number and then there's a chance and the answer would be c.72

4 0

3 years ago