MATHEMATICS MIDDLE SCHOOL

the common ratio of a geometric series is 3 and the sum of the first 8 terms is 3280. what is the first term of the series?

Answers

Answer 1

Answer:

Answer: 1

Step-by-step explanation:

The formula for calculating the sum of a Geometric series if the common ratio is greater than 1 is given as :

$S_(n)$ = $(a(r^(n)-1) )/(r-1)$

Where $S_(n)$ is the sum of terms , a is the first term , r is the common ratio and n is the number of terms.

From the question:

$S_(n)$ = 3280

a = ?

r = 3

n = 8

Substituting this into the formula , we have

3280 = $(a(3^(8)-1) )/(3-1)$

3280 = $(a(6561-1))/(2)$

Multiply through by 2 , we have

6560 = a ( 6560)

divide through by 6560, we have

6560/6560 = a(6560)/6560

Therefore : a = 1

The first term of the series is thus 1

Answer 2

Answer:

Answer:

Step-by-step explanation:

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How many times can 8 go into 60

Answers

8 Times 7 Is 56 . So That's Your Answer. You Can Not Add 8More To 56 Without Going Over 60

8 goes into 60 7 times but the answer comes out to be 7,5

Evaluate.
17 + 330 • 20 ÷ 10² ÷ 2 – 14

Answers

36
you can put it in the calculator ten squared as 10^2

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Find the x- and y- intercept of the line -3x+9y=18

Answers

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Emma's tea shop has caffeinated tea and decaffeinated tea. The tea shop swerved 30 caffeinated teas and 10 decaffeinated teas. What percentage of the teas were caffeinated?

Answers

Answer:

75% of the tea is caffeinated

Step-by-step explanation:

100 = 25 25 25 25

40 = 10 10 10 10 = 75 25

therefore if 40 = 100% then subtract 25% from 40 and you get 30 or 75%

Answer:

75% HAS CAFINEEEEE

Step-by-step explanation:

Which expression is equivalent to (-2)(a+6)?a.) -2a+6
b.) 2a+12

c.) -2a-12

d.) -2a+12

Answers

$(-2)(a+6) =-2a -12 \n \n Answer : \ c.) -2a-12$

Answer:

Step-by-step explanation:

Can someone explain to me how to do this? :

8^2y+4 = 16^y+1

Answers

Answer:

y = -4

Step-by-step explanation:

Assuming the problem is $8^(2y+4)=16^(y+1)$ , it would be nice if we could convert both sides of this equation to the same base; that way, we could compare the exponents directly in an equation of their own. Fortunately, 8 and 16 are both powers of 2 -- $2^3$ and $2^4$ , we can rewrite the original equation by substituting those in:

$(2^3)^(2y+4)=(2^4)^(y+1)$

When you have an exponent raised to another exponent, you multiply those exponents together, so we can simplify our equation by distributing a 3 in the left exponent and a 4 in the right:

$2^(3\cdot(2y+4))=2^(4\cdot(y+1))\n2^(3\cdot2y+3\cdot4)=2^(4\cdot y+4\cdot1)\n2^(6y+12)=2^(4y+4)$

With both of our bases the same, we can now simply compare their exponents directly to solve for y:

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