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Alex73 [517]
2 years ago
10

Oliver is playing a math game. He chose three cards. The value of each card is shown.

Mathematics
1 answer:
stepan [7]2 years ago
4 0

Answer:

1234567dgidchtdcjtdjutchitcbjtxvutx

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What is the sum of the geometric series 2^0 + 2^1 + 2^2 + 2^3 + 2^3 + 2^4 + … + 2^9?
GREYUIT [131]
Sum is
S_{n}=\frac{a_{1}(1-r^{n})}{1-r}

r=common ratio
a1=first term
it looks like 2^0=1 is the first term aka a1
it goes to the 9th term (2^9)

sub
S_{9}=\frac{1(1-(2)^{9})}{1-2}
S_{9}=\frac{1-512}{-1}
S_{9}=\frac{-511}{-1}
S_{9}=511

4 0
3 years ago
Read 2 more answers
Can the sum or multiplication of two palindromic numbers be another palindromic number?​
rosijanka [135]

Answer: No.

Step-by-step explanation:

If you multiply 1001 by 1001 it's not another palindromic number.

6 0
2 years ago
I do not know this answer
ch4aika [34]

Answer:

K=(3,6)

Step-by-step explanation:

If you reflect it over the y-axis the first number will change to a positive and the second number will stay the same.

Hope this helps have a great day! :)

7 0
2 years ago
At Ralph's fruitstand three apples cost of $.90 you want to buy seven apples how much will they cost
Nata [24]
1 apple is $.30. $.30 times 7 = $2.10. 
3 0
2 years ago
Prove this qns plzz ​
artcher [175]

Answer:

L.H.S.

= (cos5a.sin2a-cos4a.sin3a)/ (sin5a.sin2a-cos4a.cos3a)

Multiply numerator and denominator by 2.

= 2(cos5a.sin2a - cos4a.sin3a) / 2(sin5a.sin2a - cos4a.cos3a)

= (2cos5a.sin2a - 2cos4a.sin3a)/

(2sin5a.sin2a - 2cos4a.cos3a) = [sin(5a+2a)-sin(5a-2a)-sin(4a+3a)

+sin(4a-3a)]/[cos(5a-2a)-cos(5a+2a)-sin(4a-3a) +cos(4a+3a)]

= (sina - sin3a)/(cso3a-cosa)

= (-2cos2a.sina)/(-2sin2a.sina)

= cos2a/sin2a

= cot2a

= R.H.S.

6 0
2 years ago
Read 2 more answers
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