Answer:
I don't understand is there any thing to fill in the blanks or something?
If the third term of the aritmetic sequence is 126 and sixty fourth term is 3725 then the first term is 8.
Given the third term of the aritmetic sequence is 126 and sixty fourth term is 3725.
We are required to find the first term of the arithmetic sequence.
Arithmetic sequence is a series in which all the terms have equal difference.
Nth term of an AP=a+(n-1)d
=a+(3-1)d
126=a+2d--------1
=a+(64-1)d
3725=a+63d------2
Subtract second equation from first equation.
a+2d-a-63d=126-3725
-61d=-3599
d=59
Put the value of d in 1 to get the value of a.
a+2d=126
a+2*59=126
a+118=126
a=126-118
a=8
=a+(1-1)d
=8+0*59
=8
Hence if the third term of the arithmetic sequence is 126 and sixty fourth term is 3725 then the first term is 8.
Learn more about arithmetic progression at brainly.com/question/6561461
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D 59. Simple use of order of operations were needed.
Answer:
1 in 10 chance
Step-by-step explanation:
There are 10 cards, so since you reach into the bag and pull out only 1 card, there is a 1 in 10 chance.
Answer:
The value of cos 0 = - 2 sqrt(85) / 85 = - 0.216930458
Step-by-step explanation:
cos 0 = ?
Point on the terminal side of 0 : (-2,9)=(x,y)→x=-2, y=9
cos 0 = x/r
r=sqrt(x^2+y^2)
Replacing the known values:
r=sqrt( (-2)^2+(9)^2)
r=sqrt(4+81)
r=sqrt(85)
cos 0 = x/r
cos 0 = -2 / sqrt(85) = -2/9.219544457 = -0.216930458
Rationalizing:
cos 0 = -[2 / sqrt(85)] [sqrt(85) / sqrt(85)]
cos 0 = - 2 sqrt(85) / [sqrt(85)]^2
cos 0 = - 2 sqrt(85) / 85