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Alecsey [184]
2 years ago
7

write the explicit rule for the nth term of the arithmetic sequence. then find the 43rd term (3,5,7….)

Mathematics
1 answer:
alexira [117]2 years ago
6 0
The nth term is 2n +1 and the 43rd term is 87
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Find all solutions in the interval [0,2pi).<br><br> cos 2x + sqrt(2) sinx=1
kykrilka [37]
<span>cos 2x + sqrt(2) sinx=1
</span><span>
Note that: cos 2x = cos^2x - sin^2x = (1-sin^2x) - sin^2x = 1 - 2sin^2x.
So, when alternatively written, you have the following equation:

</span>- 2sin^2x + sqrt(2)sinx + 1 = 1
- 2sin^2x + sqrt(2)sinx = 0

Then, let z=sin(x). So you get,

- 2z^2 + sqrt(2)z = 0
z(- 2z + sqrt(2)) = 0

Either z=0, or - 2z + sqrt(2) = 0 --->  z=sqrt(2)/2.
Then, since z=0 or z=sqrt(2)/2, therefore sin(x)=0, or sin(x)=sqrt(2)/2.

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3 years ago
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prisoha [69]

Answer:

3

Step-by-step explanation:

8 0
4 years ago
What is 63% of 130?<br><br> I need it worked out pls
ahrayia [7]

Answer:

81.9

Step-by-step explanation:

63 of 130 is 48.46%

Steps to solve "what percent is 63 of 130?"

63 of 130 can be written as:

63

130

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Graph the functions on the same coordinate axis. {f(x)=−2x+1g(x)=x2−2x−3
NikAS [45]

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(2,-3) and (-2,5)

Step-by-step explanation:

Let us graph the two equations one by one.

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If we compare this equation with the slope intercept form of a line which is given as

y=mx+c

we see that m = -1 and c =1

Hence the slope of the line is -2 and the y intercept is 1. Hence one point through which it is passing is (0,1) .

Let us find another point by putting x = 1 and solving it for y

y=-2(1)+1

y=-2+1 = -1

Let us find another point by putting x = 2 and solving it for y

y=-2(2)+1

y=-4+1 = -3

Hence the another point will be (2,-3)

Let us find another point by putting x = -2 and solving it for y

y=-2(-2)+1

y=+1 = 5

Hence the another point will be (-2,5)

Now we have two points (0,1) ,(1,-1) ,  (2,-3) and (-2,5) we joint them on line to obtain our line  

2.

g(x)=y=x^2-2x-3

y=x^2-2x+1-1-3

y=(x-1)^2-4

(y+4)=(x-1)^2

It represents the parabola opening upward with vertices (1,-4)

Let us mark few coordinates so that we may graph the parabola.

i) x=0 ; y=y=(0)^2-2(0)-3=0-0-3=-3 ; (0,-3)

ii)x=-1 ; y=(-1)^2-2(-1)-3=1+2-3=0 ; (-1,0)

iii) x=2 ; y=(2)^2-2(2)-3 = 4-4-3 =-3 ;(2,-3)

iii) x=1 ; y=(1)^2-2(1)-3 = 1-2-3 =-4  ;(1,-4)

iii) x=-2 ; y=(-2)^2-2(-2)-3 = 4+4-3 =5  ;(-2,5)

Now we plot them on coordinate axis and line them to form our parabola

When we plot them we see that we have two coordinates (2,-3) and (-2,5) are common , on which our graphs are intersecting. These coordinates are solution to the two graphs.

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4 years ago
Solve compound inequality x ≥ -3 or x &lt; 3
Romashka-Z-Leto [24]

Answer:

Step-by-step explanation:

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