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2 years ago

13

A truck is bought at the price of $67,000. The price decreases at the rate of .30 every year (t). Which equation matches this ex

ponential model?

2 answers:

melamori03 [73]2 years ago

7 0

The answer is y=-3/10t+67,000

dangina [55]2 years ago

3 0

Answer:

f(t)=67,000(.70)t

Step-by-step explanation:

correct answer on my test

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What is the solution to 31|-3x + 9|=-18

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Answer:

No solutions

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3 years ago

What is the slope of the line that passes through the points (2,3) and (8,6)

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Answer:

3/6

Step-by-step explanation:

slope is rise over run so y-axis is rise x-axis is run. that gives me the answer

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3 years ago

Which equation represents a vertical line ?<br> y = -x <br> y = 12 <br> x = y <br> x = 12

OLEGan [10]

Y is a vertical axis
x is a horizontal axis
when drawing a line, the line drawn to them is perpendicular so lines only touching the x axis are vertical and lines touching only the y axis are horizontal

the
a vertical line is x=value

the only thing that works is x=12

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3 years ago

Read 2 more answers

If the revenue for the week is $2000, and labor cost consist of two workers earning $8 per hour who works 40 hours each, whay is

SVEN [57.7K]

Two workers earning 8 bucks an hour for 40hrs, well, that means each worker is making 8*40 or 320 bucks, now, is two workers, so 320+320 or 640 is the total cost then.

now, the revenue, sales income, is 2,000 bucks, how much is 640 off of?

well, if we take 2000 to be the 100%, what is 640 in percentage off of it?

$\bf \begin{array}{ccllll}
amount&\%\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
2000&100\\
640&x
\end{array}\implies \cfrac{2000}{640}=\cfrac{100}{x}$

solve for "x".

6 0

3 years ago

1000 was invested at 5.5% interest, compounded annually. After sometime the amount had grown to 1550. How long was the money inv

scZoUnD [109]

compound interest equation for annually compounded

$A=P(1+r)^t$

A=final amount

P=principal

r=rate in decimal

t=time in years

given that

A=1550

P=1000

r=5.5%=0.055

find t

$1550=1000(1+0.055)^t$

divide both sides by 1000

$1.55=1.055^t$

take ln of both sides

$ln(1.55)=ln(1.055^t)$

use ln rule $ln(a^b)=b(ln(a))$

$ln(1.55)=t(ln(1.055))$

divide both sides by ln(1.055)

$\frac{ln(1.55)}{ln(1.055)}=t$

using a calculator, we get that t=8.18544 yrs

so about 8.2yrs

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3 years ago