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3 years ago

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Choose the correct answer. Your bill from Acme is marked "3/12, net 30." The amount of the bill is $98.65. If you pay the bill w

ithin 12 days, how much should you tender?
a 95.69
b 96.57
c 97.44

2 answers:

xenn [34]3 years ago

5 0

I think it's B????? I may be wrong tho

r-ruslan [8.4K]3 years ago

4 0

Answer:

95.69

Step-by-step explanation:

this is the correct answer

brainlist if you woulds so kind

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Y varies directly with X if y equals 39 when x equals 3 What is the value of y when x equals 2

igomit [66]

Answer:

y = 26 when x = 2

Step-by-step explanation:

Y varies directly with X

y= kx

39 = 3k

k = 13

y = 13x

y(2)= 13(2) = 26

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3 years ago

A bag contains white marbles and green marbles, 62 in total. The number of white marbles is 3 less than 4 times the number of gr

Sedaia [141]

Answer:

There are 49 white marbles.

Step-by-step explanation:

w + g = 62 (w = white and g = green marbles).

w = 4g - 3

w - 4g = -3 Subtract this equation from the first equation:

g - (-4g) = 62 - (-3)

5g = 65

g = 65/5 = 13.

So w = 62 - 13 = 49.

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3 years ago

The surface of each crystal has an area of

Umnica [9.8K]

Answer:

10,000,000,000 × 10^-1, 1/1,000,000,000^-1

Step-by-step explanation:

Im not sure what you are asking

7 0

3 years ago

1-What is the sum of the series? ∑j=152j Enter your answer in the box.

tangare [24]

Answer:

Please see the Step-by-step explanation for the answers

Step-by-step explanation:

1)

∑ $\left \ {{5} \atop {j=1}} \right.$ 2j

The sum of series from j=1 to j=5 is:

∑ = 2(1) + 2(2) + 2(3) + 2(4) + 2(5)

= 2 + 4 + 6 + 8 + 10

∑ = 30

2)

This question is not given clearly so i assume the following series that will give you an idea how to solve this:

∑ $\left \ {{4} \atop {k=1}} \right.$ 2k²

The sum of series from k=1 to j=4 is:

∑ = 2(1)² + 2(2)² + 2(3)² + 2(4)²

= 2(1) + 2(4) + 2(9) + 2(16)

= 2 + 8 + 18 + 32

∑ = 60

∑ $\left \ {{4} \atop {k=1}} \right.$ (2k)²

∑ = (2*1)² + (2*2)² + (2*3)² + (2*4)²

= (2)² + (4)² + (6)² + (8)²

= 4 + 16 + 36 + 64

∑ = 120

∑ $\left \ {{4} \atop {k=1}} \right.$ (2k)²- 4

∑ = (2*1)²-4 + (2*2)²-4 + (2*3)²-4 + (2*4)²-4

= (2)²-4 + (4)²-4 + (6)²-4 + (8)²-4

= (4-4) + (16-4) + (36-4) + (64-4)

= 0 + 12 + 32 + 60

∑ = 104

∑ $\left \ {{4} \atop {k=1}} \right.$ 2k²- 4

∑ = 2(1)²-4 + 2(2)²-4 + 2(3)²-4 + 2(4)²-4

= 2(1)-4 + 2(4)-4 + 2(9)-4 + 2(16)-4

= (2-4) + (8-4) + (18-4) + (32-4)

= -2 + 4 + 14 + 28

∑ = 44

3)

∑ $\left \ {{6} \atop {k=3}} \right.$ (2k-10)

∑ = (2×3−10) + (2×4−10) + (2×5−10) + (2×6−10)

= (6-10) + (8-10) + (10-10) + (12-10)

= -4 + -2 + 0 + 2

∑ = -4

4)

1+1/2+1/4+1/8+1/16+1/32+1/64

This is a geometric sequence where first term is 1 and the common ratio is 1/2 So

a = 1

This can be derived as

1/2/1 = 1/2 * 1 = 1/2

1/4/1/2 = 1/4 * 2/1 = 1/2

1/8/1/4 = 1/8 * 4/1 = 1/2

1/16/1/8 = 1/16 * 8/1 = 1/2

1/32/1/16 = 1/32 * 16/1 = 1/2

1/64/1/32 = 1/64 * 32/1 = 1/2

Hence the common ratio is r = 1/2

So n-th term is:

$ar^{n-1}$ = $1(\frac{1}{2})^{n-1}$

So the answer that represents the series in sigma notation is:

∑ $\left \ {{7} \atop {j=1}} \right.$ $(\frac{1}{2})^{j-1}$

5)

−3+(−1)+1+3+5

This is an arithmetic sequence where the first term is -3 and the common difference is 2. So

a = 1

This can be derived as

-1 - (-3) = -1 + 3 = 2

1 - (-1) = 1 + 1 = 2

3 - 1 = 2

5 - 3 = 2

Hence the common difference d = 2

The nth term is:

a + (n - 1) d

= -3 + (n−1)2

= -3 + 2(n−1)

= -3 + 2n - 2

= 2n - 5

So the answer that represents the series in sigma notation is:

∑ $\left \ {{5} \atop {j=1}} \right.$ (2j−5)

6 0

3 years ago

What is the relationship between the exponents of (3^2) ^4 and the exponents of 3 ^8

zysi [14]

Answer:

They are equal

Step-by-step explanation:

Using the rule of exponents

$(a^m)^{n}$ = $a^{mn}$

Given

$(3^2)^{4}$ = $3^{2.4}$ = $3^{8}$

The exponents of both expression are 8 and therefore equal

4 0

3 years ago