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julsineya [31]
2 years ago
7

PLEASE HELP ASAP, WILL MARK BRAINLIEST

Mathematics
1 answer:
Dahasolnce [82]2 years ago
4 0

Answer:

number  1. C^2 if C>0    number 2. Q^-5 if Q<0.      number 3. G^3 if G>0.

number 4. P^-3 if P<0

Step-by-step explanation:

hope this helps .-.

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The area of a circle is 16pi cm2. What is the circle's circumstance?
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Find the sum of each series, if it exists<br>91 + 85 + 79 + … + (­29)
Natali [406]

Answer:

651.

Step-by-step explanation:

Note: In the given series it should be -29 instead of 29 because 29 cannot be a term of AP whose first term is 91 and common difference is -6.

Consider the given series is

91+85+79+...+(-29)

It is the sum of an AP. Here,

First term = 91

Common difference = 85 - 91 = -6

Last term = -29

nth term of an AP is

a_n=a+(n-1)d

where, a is first term and d is common difference.

-29=91+(n-1)(-6)

-29-91=(n-1)(-6)

\dfrac{-120}{-6}=(n-1)

20=(n-1)

n=20+1=21

Sum of AP is

Sum=\dfrac{n}{2}[\text{First term + Last term}]

Sum=\dfrac{21}{2}[91+(-29)]

Sum=\dfrac{21}{2}[62]

Sum=651

Therefore, the sum of given series is 651.

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