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3 years ago

Find the nth term of 5 9 14 18 5 23

1 answer:

8 0

first of all it is not in the form of A. P because d=a2-a1=9-5=4,a3-a2=14-9=4,a4-a3=18-14=4,a5-a4=5-18= -13,a6-a5=23-5=18

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M=1.33333(N+20). If N=16,find M

aalyn [17]

Answer:

M=47.99988

Step-by-step explanation:

M=1.33333(16+20)

=1.33333(36)

=47.99988

8 0

3 years ago

Describe the steps that you would use to generate an expression equivalent to 8.1b+6.7a+2.5+7+.9a-2.8b

bija089 [108]

Explanation:

1. Identify the different constellations of variables. Here there are three:

b
a
(none)

2. Combine coefficients of each of the different variable constellations:

(8.1 -2.8)b +(6.7 +0.9)a +(2.5 +7)

5.3b +7.8a +9.5

3. Perform any other operations that might be required depending on the sort of equivalent wanted. For example, one could write ...

5.3(b +(78/53)a) +9.5 . . . . . . . shows the weight of a relative to b

6 0

3 years ago

Select the correct answer.

zepelin [54]

Answer:

Answer is option c

Step-by-step explanation:

$given \: equations \: are \\ x - 2y = 15 \: \: and \: 2x + 4y = - 18 \\ divide \: by \: 2 \: in \: second \: equation \\ x + 2y = - 9 \: \: (equation \: 3) \\ adding \: 1and \: 3 \: we \: get \\ 2x = 6 \\ x = 3 \\ substitute \: x = 3 \: in \: equation \: 3 \\ 3 + 2y = - 9 \\ 2y = - 9 - 3 \\ 2y = - 12 \\ y = - 6$

x=3 and y=-6

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THANKS FOR GIVING ME THE OPPORTUNITY TO ANSWER YOUR QUESTION. 

6 0

3 years ago

Read 2 more answers

What's a unknown point thats y-coordinate is zero.

klemol [59]

Answer:

an unknown point? like 0,7 or 0,5 or 0,1 or 0,0 or 0,anything?

Step-by-step explanation:

4 0

2 years ago

Find [5(cos 330 degrees + I sin 330 degrees)]^3

earnstyle [38]

Given a complex number in the form:
$z= \rho [\cos \theta + i \sin \theta]$
The nth-power of this number, $z^n$ , can be calculated as follows:

- the modulus of $z^n$ is equal to the nth-power of the modulus of z, while the angle of $z^n$ is equal to n multiplied the angle of z, so:
$z^n = \rho^n [\cos n\theta + i \sin n\theta ]$
In our case, n=3, so $z^3$ is equal to
$z^3 = \rho^3 [\cos 3 \theta + i \sin 3 \theta ] = (5^3) [\cos (3 \cdot 330^{\circ}) + i \sin (3 \cdot 330^{\circ}) ]$ (1)
And since
$3 \cdot 330^{\circ} = 990^{\circ} = 2\pi +270^{\circ}$
and both sine and cosine are periodic in $2 \pi$ , (1) becomes
$z^3 = 125 [\cos 270^{\circ} + i \sin 270^{\circ} ]$

6 0

3 years ago