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kirill [66]
1 year ago
7

A rental car company charges $68.64 per day to rent a car and $0.12 for every mile driven. Jaxon wants to rent a car, knowing th

at: • He plans to drive 500 miles. • He has at most $210 to spend. What is the maximum number of days that Jaxon can rent the car while staying within his budget?​
Mathematics
1 answer:
attashe74 [19]1 year ago
3 0

Answer:

2 days

Step-by-step explanation:

According to the problem, calculations of given data are as follows,

Rental charges per day = $68.64

Rental per mile = $0.12

Total miles = 500

Total spend = $210

Let number of days = X

So, we can find number of days by using following equation,

($68.64 × X ) + ( $0.12 × 500) = $210

$68.64 X + $60 = $210

$68.64 X = $210 - $60

$68.64 X = $150

X = $150 ÷ $68.64

X = 2.18 or 2 Days

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Tyler needs to complete the table for his consumer science class. He knows that 1 tablespoon contains 3 teaspoons and that 1 cup
olga_2 [115]

Answer:

1 cup = 48 teaspoons

C = 48t

Step-by-step explanation:

Given:

1 tablespoon = 3 teaspoons

1 cup = 16 tablespoon

Write an equation that represents the number of teaspoons, t, contained in a cup, C.​

Since

1 tablespoon = 3 teaspoons

And

1 cup = 16 tablespoon

Then

1 cup = 3 teaspoons × 16

1 cup = 48 teaspoons

Let

cup = C

Teaspoons = t

Therefore,

C = 48t

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2 years ago
Period 1 has 15 students in it and a test average of 86%. Period 2 has 21 students in it and an average of 88%. Period 3 has 12
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Add all of the percents up and then divide by 3. You get 89.6% 
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2 years ago
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Paul opened a bakery. The net value of the bakery (in
Umnica [9.8K]

Answer:

1) 2(t - 7)(t + 1)

2) $32,000

Step-by-step explanation:

v(t) = 2t² - 12t - 14

v(t) = 2(t² - 6t - 7) = 2(t - 7)(t + 1)

x = -b/2a = 6/2 = 3

y = 2(3)² - 12(3) - 14 = -32

(3, -32) means at 3 months, -32 thousand dollars

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2 years ago
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G identify the solution of the recurrence relation an = 6an − 1 – 8an − 2 for n ≥ 2 together with the initial conditions a0 = 4,
maksim [4K]
Via the generating function method, let

G(x)=\displaystyle\sum_{n\ge0}a_nx^n

Then take the recurrence,

a_n=6a_{n-1}-8a_{n-2}

multiply everything by x^n and sum over all n\ge2:

\displaystyle\sum_{n\ge2}a_nx^n=6\sum_{n\ge2}a_{n-1}x^n-8\sum_{n\ge2}a_{n-2}x^n

Re-index the sums or add/remove terms as needed in order to be able to express them in terms of G(x):

\displaystyle\sum_{n\ge2}a_nx^n=\sum_{n\ge0}a_nx^n-(a_0-a_1x)=G(x)-4-10x

\displaystyle\sum_{n\ge2}a_{n-1}x^n=\sum_{n\ge1}a_nx^{n+1}=x\sum_{n\ge1}a_nx^n=x\left(G(x)-a_0\right)=x(G(x)-4)

\displaystyle\sum_{n\ge2}a_{n-2}x^n=\sum_{n\ge0}a_nx^{n+2}=x^2\sum_{n\ge0}a_nx^n=x^2G(x)

So the recurrence relation is transformed to

G(x)-4-10x=6x(G(x)-4)-8x^2G(x)
(1-6x+8x^2)G(x)=4-14x
G(x)=\dfrac{4-14x}{1-6x+8x^2}=\dfrac{4-14x}{(1-4x)(1-2x)}=\dfrac1{1-4x}+\dfrac3{1-2x}

For appropriate values of x, we can express the RHS in terms of geometric power series:

G(x)=\displaystyle\sum_{n\ge0}(4x)^n+3\sum_{n\ge0}(2x)^n=\sum_{n\ge0}\bigg(4^n+3\cdot2^n\bigg)x^n

which tells us that

a_n=4^n+3\cdot2^n
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