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3 years ago

Plss help me!!

Mathematics

1 answer:

lakkis [162]3 years ago

4 0

Step-by-step explanation:

The problem statement says λ<1. If this is true, then the cannonball is slower than the boat, and it will never reach the boat. So I assume it's actually λ>1.

If we say t is the amount of time, then the boat travels south a distance of ut, and the cannonball travels a distance of λut.

v = λut

ut = v/λ

Using Pythagorean theorem:

d² + (ut)² > v²

d² + (v/λ)² > v²

λ²d² + v² > λ²v²

λ²d² > (λ² − 1) v²

λ² / (λ² − 1) > (v/d)²

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A building is 200ft long and 140 ft wide a 1/500 model is built of the building how long and how wide is the model?

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Let's start with the length. All you need to do is multiply is by 1/500.
You could also find the decimal equivalent to 1/500.
1/500= .002
200×.002= .4
So length of the model is .4 ft.
The width:
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The width of the model is .28 ft.
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A new car is purchased for $17,000 and over time its value depreciates by one half every 3 years. What is the value of the car 1

Jet001 [13]

Answer:

The answer is "$238".

Step-by-step explanation:

Current worth $= \$ 17,000$

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time= 19 years

depreciates rate=?

Using formula:

$\to \text{Worth= Current worth}(1- \frac{\text{depreciates rate}}{100})^{time}$

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$\to A_t=A_0(1-\frac{r}{100})^t\\\\\to 8,500= 17,000(1-\frac{r}{100})^3\\\\\to \frac{8,500}{17,000}= (1-\frac{r}{100})^3\\\\\to \frac{1}{2}= (1-\frac{r}{100})^3\\\\\to (\frac{1}{2})^{\frac{1}{3}}= (1-\frac{r}{100})\\\\\to 0.793700526 = (1-\frac{r}{100})\\\\\to \frac{r}{100} = (1-0.793700526)\\\\\to \frac{r}{100} = (1-0.8)\\\\\to r= 0.2 \times 100 \\\\\to r= 20 \%$

depreciates rate= 20%

$\to \text{Worth= Current worth}(1- \frac{\text{depreciates rate}}{100})^{time}$

$= \$ 17,000 (1- \frac{20}{100})^{19}\\\\= \$ 17,000 (1-0.2)^{19}\\\\= \$ 17,000 (0.8)^{19}\\\\= \$ 17,000 \times 0.014\\\\= \$ 238$

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