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Nina [5.8K]
3 years ago
13

What are the first 5 terms of 30-n^2

Mathematics
1 answer:
sleet_krkn [62]3 years ago
6 0

Answer:

29, 26, 21, 14, 5

Step-by-step explanation:

To find the first 5 terms, substitute n = 1, 2, 3, 4, 5 into the n th term formula.

30 - 1² = 30 - 1 = 29

30 - 2² = 30 - 4 = 26

30 - 3² = 30 - 9 = 21

30 - 4² = 30 - 16 = 14

30 - 5² = 30 - 25 = 5

The first 5 terms are 29, 26, 21, 14, 5

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Arada [10]

Answer:

T = 4.0 s

Step-by-step explanation:

The period of a simple pendulum is given by

period = 2\pi \sqrt{ \frac{l}{g} }

where l = length of the pendulum

= 4 m

g = acceleration due to gravity

= 9.8 m/s^2

period = 2\pi \sqrt{ \frac{(4 \: m)}{(9.8 \:  \frac{m}{ {s}^{2} } )} }  = 4.0 \: s

8 0
3 years ago
T = LxS solve for x. whats the anwser xD
Nookie1986 [14]
I'm pretty sure the answer is T/LS=x or in other words T divided by LS equals x.
4 0
3 years ago
There once was a student named chandler. In 5 minutes, chandler texted 435 words. At the rate how many words did chandler text p
Sergio [31]

Answer:

87

Step-by-step explanation:

7 0
3 years ago
Write the trigonometric expression <br> Cos (arcsin (u))<br> As an algebraic expression in u
makvit [3.9K]
When we use arcsine, we are finding the angle while giving the trigonometric ratio.

Arcsin(u) = theta can be rewritten as:

sin(theta) = u

Sine is opposite over hypotenuse, so u/1 means that the side opposite to theta (the y value) is u, and the hypotenuse is 1.

We can use Pythagorean Theorem to find the adjacent (x value).

1^2 - u^2 = x^2

x = sqrt(1-u^2)

Back to the original question, we are trying to find cos(arcsin(u)). We just solved all the sides for our triangle using arcsin(u). Now we need to do cos(u).

Cosine is adjacent over hypotenuse.

So our answer is sqrt(1-u^2)/1

Or just sqrt(1-u^2)







5 0
3 years ago
YA
irakobra [83]

Answer:

a) P(0, <u>2</u>), Q(<u>4</u>, 0)

b) Please find attached the plot of the points P and Q on a chart made with MS Excel

c) Please find the graph of the line that represent the function 4·x + 2·y = 8 for values of x from -2 to 3 made with the Insert Chart feature on MS Excel

Step-by-step explanation:

The given equation for the line is 4·x + 2·y = 8

a) The coordinates of P = P(0, _)

Therefore, the point 'P', which is the point where the variable y = 0, is the point the (straight line) graph intercepts the x-axis (the x-intercept)

When y = 0 from the given equation, we get;

4·x + 2·y = 8

At the point y = 0;

4·x + 2 × 0 = 8

x = 8/4 = 2

x = 2

∴ The coordinates of P = P(0, _) = P(0, <u>2</u>)

Similarly, when x = 0, we get;

4·x + 2·y = 8

At the point x = 0;

4 × 0 + 2·y = 8

y = 8/2 = 4

y = 4

∴ The coordinates of Q = Q(_, 0) = Q(<u>4</u>, 0)

b) Rewriting the given equation in terms of 'y' gives;

y = (8 - 4·x)/2 = 4 - 2·x

y = 4 - 2·x

With the help of MS Excel, the points P and Q are plotted in the attached graph

c) The line of the graph of the function 4·x + 2·y = 8 for values of x from -2 to 3 can be added by Changing the Chart Type to 'Scatter with Smooth Lines and Markers' within MS Excel as presented in the included graph of the line.

3 0
2 years ago
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