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liubo4ka [24]
3 years ago
9

The cost for renting a car is $29.00 a day plus $0.29 for every kilometre after the first 100 km. what is the cost of renting a

car after 14 days and 320 km?
(linear equations)
Mathematics
1 answer:
lidiya [134]3 years ago
4 0

Answer:

$469.80

Step-by-step explanation:

First, let's find out how many fees would be paid exclusively for renting it out for 14 days.

29.00 * 14 = $406

They have to pay 406 dollars just for having the car for 14 days.

Now let's find out how much will need to paid for driving for 320 km.

We can first take away 100 km because the fee only starts afterwards.

320 - 100 = 220

220 * 0.29 = 63.8

63.80 + 406 = 469.80

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Answer:

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Step-by-step explanation:

vertex angles are congruent so you can set them equal to each other and solve

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8 0
3 years ago
Kelly deposits $3,000 at a rate of 5.2% compounded quarterly. With no additional deposits or withdrawals, what is the account ba
lorasvet [3.4K]
\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$3000\\
r=rate\to 5.2\%\to \frac{5.2}{100}\to &0.052\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{quarterly, thus four}
\end{array}\to &4\\
t=years\to &10
\end{cases}
\\\\\\
A=3000\left(1+\frac{0.052}{4}\right)^{4\cdot 10}\implies A=3000(1.013)^{40}
\\\\\\
A\approx 5029.201878874637
7 0
3 years ago
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X^2-6x+5=0 solve with quadratic formula
Yuliya22 [10]

Answer: x1  = 5

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Step-by-step explanation:

1. Identify the coefficients

2. Move the constant to the right side of the equation and combine

3.  Complete the square

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Which equation represents a line which is perpendicular to the line y=−4x−5?
AnnZ [28]

Answer:

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Step-by-step explanation:

5 0
3 years ago
2. Mourette is a member of her high school golf team. She hits a golf ball
Llana [10]

Answer:

a. h = 60t − 4.9t²

b. 12.2 seconds

c. 183.7 meters

Step-by-step explanation:

a. Given:

y₀ = 0 m

v₀ = 60 m/s

a = -9.8 m/s²

y = y₀ + v₀ t + ½ at²

h = 0 m + (60 m/s) t + ½ (-9.8 m/s²) t²

h = 60t − 4.9t²

b. When the ball lands, h = 0.

0 = 60t − 4.9t²

0 = t (60 − 4.9t)

t = 0 or 12.2

The ball lands after 12.2 seconds.

c. The maximum height is at the vertex of the parabola.

t = -b / (2a)

t = -60 / (2 × -4.9)

t = 6.1 seconds

Alternatively, the maximum height is reached at half the time it takes to land.

t = 12.2 / 2

t = 6.1 seconds

After 6.1 seconds, the height reached is:

h = 60 (6.1) − 4.9 (6.1)²

h = 183.7 meters

7 0
3 years ago
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