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OLga [1]
3 years ago
12

Chu-Li has $100 in her piggy bank. How much will she have in her bank 40 weeks from now if she puts $1 in the bank next week, $2

two weeks from now, $3 three weeks from now, and continues to increase the amount that she puts in by $1 each week?
Mathematics
1 answer:
BigorU [14]3 years ago
6 0

Answer:

  $920

Step-by-step explanation:

The amounts added are the arithmetic sequence ...

  1, 2, 3, ... 40

The average term of this sequence is (1 +40)/2 = 20.5. The sum of the terms of this sequence is the number of terms (40) multiplied by the average term (20.5). That product is ...

  40·20.5 = 820

Chu-Li will have added $820 to her bank in 40 weeks, so the total there is ...

  $100 +820 = $920 . . . amount Chu-Li will have 40 weeks from now

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4 years ago
Dan buys a car for £2200.
KatRina [158]

Answer:

The worth of the car after 6 years is £2,134.82

Step-by-step explanation:

The amount at which Dan buys the car, PV = £2200

The rate at which the car depreciates, r = -0.5%

The car's worth, 'FV', in 6 years is given as follows;

FV = PV \cdot \left ( 1 + \dfrac{r}{100} \right )^n

Where;

r = The depreciation rate (negative) = -0.5%

FV = The future value of the asset

PV = The present value pf the asset = £2200

n = The number of years (depreciating) = 6

By plugging in the values, we get;

FV = 2200 \times \left ( 1 + \dfrac{-0.5}{100} \right )^6 \approx 2,134.82

The amount the car will be worth which is its future value, FV after 6 years is FV ≈ £2,134.82 (after rounding to the nearest penny (hundredth))

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Answer:

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Step-by-step explanation:

6 0
3 years ago
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The volume V of an ice cream cone is given by V = 2 3 πR3 + 1 3 πR2h where R is the common radius of the spherical cap and the c
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Answer:

The change in volume is estimated to be 17.20 \rm{in^3}

Step-by-step explanation:

The linearization or linear approximation of a function f(x) is given by:

f(x_0+dx) \approx f(x_0) + df(x)|_{x_0} where df is the total differential of the function evaluated in the given point.

For the given function, the linearization is:

V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh

Taking R_0=1.5 inches and h=3 inches and evaluating the partial derivatives we obtain:

V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh\\V(R, h) = V(R_0, h_0) + (\frac{2 h \pi r}{3}  + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh

substituting the values and taking dx=0.1 and dh=0.3 inches we have:

V(R_0+dR, h_0+dh) =V(R_0, h_0) + (\frac{2 h \pi r}{3}  + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh\\V(1.5+0.1, 3+0.3) =V(1.5, 3) + (\frac{2 \cdot 3 \pi \cdot 1.5}{3}  + 2 \pi 1.5^2)\cdot 0.1 + (\frac{\pi 1.5^2}{3} )\cdot 0.3\\V(1.5+0.1, 3+0.3) = 17.2002\\\boxed{V(1.5+0.1, 3+0.3) \approx 17.20}

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