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3 years ago

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Mathematics

2 answers:

garri49 [273]3 years ago

7 0

Answer:

120 markers

Step-by-step explanation:

I'm pretty sure this isn't the way to solve this...but it works

8+5=13

195/13=15

15*8=120

Andru [333]3 years ago

3 0

120
Yuppsssssssssssssssssss

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Jonathan bought a Gucci belt in Chinatown for $35.00. The belt was on sale for ⅓ off the regular price. Then, an additional 8% w

leonid [27]

Answer:

<u>The original price of the belt was $ 59.32 and Jonathan paid $ 35 for it.</u>

Step-by-step explanation:

1. Let's review the information given to us to answer the question correctly:

Amount Jonathan paid for the belt = $ 35

Discount = 1/3 off the regular price

Coupon = additional 8% = 0.08

2. How much did the Gucci belt cost Jonathan? Show your work or explain how you know.

We assume the question is asking for the original price of the belt, then we make this calculation:

Amount Jonathan paid for the belt = (Original price - Discount) - Coupon

Replacing with the values we know:

Original price = x

35 = (x - 1/3x) - 0.08x

35 = 2/3x - 0.08x

35 = 0.67x - 0.08x

35 = 0.59x

x = 35/0.59

x = 59.32

<u>The original price of the belt was $ 59.32 and Jonathan paid $ 35 for it.</u>

4 0

3 years ago

<img src="https://tex.z-dn.net/?f=%24a%2Ba%20r%2Ba%20r%5E%7B2%7D%2B%5Cldots%20%5Cinfty%3D15%24%24a%5E%7B2%7D%2B%28a%20r%29%5E%7B

riadik2000 [5.3K]

Let

$S_n = \displaystyle \sum_{k=0}^n r^k = 1 + r + r^2 + \cdots + r^n$

where we assume |r| < 1. Multiplying on both sides by r gives

$r S_n = \displaystyle \sum_{k=0}^n r^{k+1} = r + r^2 + r^3 + \cdots + r^{n+1}$

and subtracting this from $S_n$ gives

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

As n → ∞, the exponential term will converge to 0, and the partial sums $S_n$ will converge to

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

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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