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3 years ago

Which of the following does not have a scale factor of 1/60? (Hint: Use Unit Conversion).

1 answer:

4 0

1in : 5ft
1in : (5 x 12) in
1in : 60 in

1ft : 20yd
1ft : (20 x 3) ft
1ft : 60 ft

1 cm : 60cm

1in : 60ft
1in : (60 x 12)in
1in : 720 in
Hence, 1in : 60ft does not have a scale factor of 1/60

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The value V of a certain automobile that is t years old can be modeled by V(t) = 14,651(0.81). According to the model, when will

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Answer:

a) The car will be worth $8000 after 2.9 years.

b) The car will be worth $6000 after 4.2 years.

c) The car will be worth $1000 after 12.7 years.

Step-by-step explanation:

The value of the car after t years is given by:

$V(t) = 14651(0.81)^{t}$

According to the model, when will the car be worth V(t)?

We have to find t for the given value of V(t). So

$V(t) = 14651(0.81)^{t}$

$(0.81)^t = \frac{V(t)}{14651}$

$\log{(0.81)^{t}} = \log{(\frac{V(t)}{14651})}$

$t\log{(0.81)} = \log{(\frac{V(t)}{14651})}$

$t = \frac{\log{(\frac{V(t)}{14651})}}{\log{0.81}}$

(a) $8000

V(t) = 8000

$t = \frac{\log{(\frac{8000}{14651})}}{\log{0.81}} = 2.9$

The car will be worth $8000 after 2.9 years.

(b) $6000

V(t) = 6000

$t = \frac{\log{(\frac{6000}{14651})}}{\log{0.81}} = 4.2$

The car will be worth $6000 after 4.2 years.

(c) $1000

V(t) = 1000

$t = \frac{\log{(\frac{1000}{14651})}}{\log{0.81}} = 12.7$

The car will be worth $1000 after 12.7 years.

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