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Ne4ueva [31]
3 years ago
10

Benito spent $1616 to operate his car last year. Some of these expenses are listed below. Benito's only other expense was for ga

soline. If he drove 7270 miles, what was the average cost of the gasoline per mile?
Operating Expenses
Insurance $946
Registration $105
Maintenance $74
A. 6.75 cents
B. 0.81 cents
C. 62.30 cents
D. not enough information given
Mathematics
1 answer:
Verizon [17]3 years ago
5 0
Subtract all the operating expenses from the amount he spent total and get what's left over is $491.  That's what he spent on gas.  Divide the number of miles he drove by the amount he spent on the gas to find out what the average cost of the gas was per mile.  491/7270 = .06753.  But that is expressed in dollars and we need it in cents.  Multiply that decimal by 100 to get 6.75 cents, choice A from above.
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I don’t know the answer
anastassius [24]
I THINK IT IS B

B) D(X)= X-8x4 
7 0
3 years ago
Order these numbers from least to greatest.<br> 2<br> 12/5<br> 2.47<br> square root of 3
Arturiano [62]

Answer:

square root, 2,12/5,2.47

Step-by-step explanation:

7 0
3 years ago
P=3w^2. <br> Calculate P when w=4.
algol13
P= 3(4)^2
P= 12^2
P = 144
4 0
3 years ago
How many video games can you buy with $125 if one video game costs $5
GalinKa [24]

Your answer's 25

You have to divide 125 times 5 to get your answer of 25

<h2><u>Hope this helps</u></h2>
5 0
1 year ago
Read 2 more answers
List all possible rational roots. Then use synthetic division to confirm which rational roots exist:
Kisachek [45]

Answer:

\boxed{(1) \, x = \, \pm \dfrac{1}{2}, \pm 1, \pm2, \pm \dfrac{5}{2}, \pm 5, \pm 10; (2) \, x = -2}

Step-by-step explanation:

2x³+ 6x² - x - 10 = 0

(1) Possible roots

The Rational Roots Theorem states that, if a polynomial has any rational roots, they will have the form p/q, where p is a factor of the constant term  and q is a factor of the leading coefficient.

\text{Possible rational root} = \dfrac{ p }{ q } = \dfrac{\text{factor of constant term}}{\text{factor of leading coefficient}}

In your function, the constant term is -10 and the leading coefficient is 2, so

\text{Possible root} = \dfrac{\text{factor of 10}}{\text{factor of 2}}

Factors of 10 = ±1, ±2, ±5, ±10

Factors of 2 = ±1, ±2

\text{Possible roots are } \large \boxed{\mathbf{x = \pm \dfrac{1}{2}, \pm 1, \pm2, \pm \dfrac{5}{2}, \pm 5, \pm 10}}

(2) Synthetic division

Rather than work through all 12 possibilities, I will do one that works.

\begin{array}{r|rrrr}-2 & 2 & 6 & -1 & -10\\& & -4& -4 & 10\\& 2 & 2& -5 & 0\\\end{array}

So, x = -2 is a root, and the quotient is 2x² + 2x - 5.

(3) Check for other rational roots

2x² + 2x - 5 = 0

D = b² - 4ac =2²- 4(2)(-5) = 4 + 40 = 44

√44 = 2√11, which is irrational.

Since irrational roots come in pairs, the cubic equation has two real, irrational roots and one rational root at x = -2.

6 0
3 years ago
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