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3 years ago

A......B......C.......or D

Mathematics

2 answers:

Dima020 [189]3 years ago

8 0

A is the right answer

IrinaVladis [17]3 years ago

5 0

I think the answe is a

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PLEASE HELP !!!!!!! GEOMETRY

nata0808 [166]

I came up with d and e as an answer

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3 years ago

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

lakkis [162]

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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2 years ago

What function/equation is best for this

professor190 [17]

The answer is y= -2x !

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3 years ago

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PLEASE HURRY IM TIMED!!!

andriy [413]

Answer-

A. Brandon’s sound intensity level is 1/4th as compared to Ahmad’s.

Solution-

Given that, loudness measured in dB is

$L=10\log \frac{I}{I_0}$

Where,

I = Sound intensity,

I₀ = 10⁻¹² and is the least intense sound a human ear can hear

Given in the question,

I₁ = Intensity at Brandon's = 10⁻¹⁰

I₂ = Intensity at Ahmad's = 10⁻⁴

Then,

$L_1=10\log \frac{I_1}{I_0}=10\log \frac{10^{-10}}{10^{-12}}=10\log \frac{1}{10^{-2}}=10\log 10^{2}=2\times10\log 10=20$

$L_2=10\log \frac{I_2}{I_0}=10\log \frac{10^{-4}}{10^{-12}}=10\log \frac{1}{10^{-8}}=10\log 10^{8}=8\times10\log 10=80$

$\therefore \frac{L_1}{L_2} =\frac{20}{80} =\frac{1}{4}$

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3 years ago

How many significant figures are represented in 0.00509?

Minchanka [31]

Answer:

there are three significant numbers

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2 years ago

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