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1 year ago

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Maggie brought oranges and sandwiches to a picnic. The number of oranges was three more than twice the number of sandwiches. Mag

gie brought 9 oranges to the picnic. How many sandwiches did she bring?

1 answer:

tatuchka [14]1 year ago

8 0

Maggie brought 3 sandwiches to the picnic .

In the question ,

it is given that

Maggie bought oranges and sandwiches to a picnic .

given the number of oranges = 9

also given that , number of oranges was three more than twice the number of sandwiches

let the number of sandwiches Maggie brought to a picnic be x .

So , the number of oranges is = 2x + 3

So , 2x + 3 = 9

Subtracting 3 from both the sides ,

we get ,

2x = 9 - 3

2x = 6

On dividing both sides by 2,

we get ,

x = 6/2

x = 3

Therefore , the number of sandwiches = 3 .

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3 years ago

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Let

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and subtracting this from $S_n$ gives

$(1 - r) S_n = 1 - r^{n+1} \implies S_n = \dfrac{1 - r^{n+1}}{1 - r}$

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$\displaystyle \lim_{n\to\infty} S_n = \dfrac1{1-r}$

Now, we're given

$a + ar + ar^2 + \cdots = 15 \implies 1 + r + r^2 + \cdots = \dfrac{15}a$

$a^2 + a^2r^2 + a^2r^4 + \cdots = 150 \implies 1 + r^2 + r^4 + \cdots = \dfrac{150}{a^2}$

We must have |r| < 1 since both sums converge, so

$\dfrac{15}a = \dfrac1{1-r}$

$\dfrac{150}{a^2} = \dfrac1{1-r^2}$

Solving for r by substitution, we have

$\dfrac{15}a = \dfrac1{1-r} \implies a = 15(1-r)$

$\dfrac{150}{225(1-r)^2} = \dfrac1{1-r^2}$

Recalling the difference of squares identity, we have

$\dfrac2{3(1-r)^2} = \dfrac1{(1-r)(1+r)}$

We've already confirmed r ≠ 1, so we can simplify this to

$\dfrac2{3(1-r)} = \dfrac1{1+r} \implies \dfrac{1-r}{1+r} = \dfrac23 \implies r = \dfrac15$

It follows that

$\dfrac a{1-r} = \dfrac a{1-\frac15} = 15 \implies a = 12$

and so the sum we want is

$ar^3 + ar^4 + ar^6 + \cdots = 15 - a - ar - ar^2 = \boxed{\dfrac3{25}}$

which doesn't appear to be either of the given answer choices. Are you sure there isn't a typo somewhere?

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Answer:

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