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3 years ago

Kendra wants to pay 500 photographs into an album 6 photos to a page she figures that she will need about 100 pages

Mathematics

2 answers:

alina1380 [7]3 years ago

8 0

Divide your photos by the number of photos

500/6=83.3333 so she would need at least 84 pgs to fit the photos

uysha [10]3 years ago

5 0

Answer:

Step-by-step explanation: 6x100=600

600 is too many pages.

500 / 6 = 83.3

She will need about 83 pages

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Someeee one?????????????????

Ad libitum [116K]

Answer:

Option B) $a_{n} = 2\cdot 4^{n-1}$

Step-by-step explanation:

The given geometric sequence is

2, 8, 32, 128,....

The general form of a geometric sequence is given by

$a_{n} = a_{1}\cdot r^{n-1}$

Where n is the nth term that we want to find out.

a₁ is the first term in the geometric sequence that is 2

r is the common ratio and can found by simply dividing any two consecutive numbers in the sequence,

$r=\frac{8}{2} = 4$

You can try other consecutive numbers too, you will get the same common ratio

$r=\frac{32}{8} = 4$

$r=\frac{128}{32} = 4$

So the common ratio is 4 in this case.

Substitute the value of a₁ and r into the above general equation

$a_{n} = 2\cdot 4^{n-1}$

This is the general form of the given geometric sequence.

Therefore, the correct option is B

Note: Don't multiply the first term and common ratio otherwise you wont get correct results.

Verification:

$a_{n} = 2\cdot 4^{n-1}$

Lets find out the 2nd term

Substitute n = 2

$a_{2} = 2\cdot 4^{2-1} = 2\cdot 4^{1} = 2\cdot 4 = 8$

Lets find out the 3rd term

Substitute n = 3

$a_{3} = 2\cdot 4^{3-1} = 2\cdot 4^{2} = 2\cdot 16 = 32$

Lets find out the 4th term

Substitute n = 4

$a_{4} = 2\cdot 4^{4-1} = 2\cdot 4^{3} = 2\cdot 64 = 128$

Lets find out the 5th term

Substitute n = 5

$a_{5} = 2\cdot 4^{5-1} = 2\cdot 4^{4} = 2\cdot 256 = 512$

Hence, we are getting correct results!

6 0

3 years ago

Please help me I will give a brainly!!!!

nydimaria [60]

Answer:

Yes

Step-by-step explanation:

1 : The athlete's hands push the medicine ball forward. The medicine ball pushes the athlete's hands backward.

2: Friction

3: The first pair of action-reaction force pairs is: foot A pushes ball B to the right; and ball B pushes foot A to the left. The second pair of action-reaction force pairs is: foot C pushes ball B to the left; and ball B pushes foot C to the right

7 0

2 years ago

A^-1/[a^-1 - b^-1] + a^-1/[a^-1 + b^-1]

o-na [289]

$\dfrac{a^{-1}}{a^{-1}-b^{-1}}+\dfrac{a^{-1}}{a^{-1}+b^{-1}}=a^{-1}\left(\dfrac{a^{-1}+b^{-1}}{(a^{-1}-b^{-1})(a^{-1}+b^{-1})}+\dfrac{a^{-1}-b^{-1}}{(a^{-1}+b^{-1})(a^{-1}-b^{-1})}\right)$