All categories

3 years ago

Cabs R’ Us charges $2.00 plus an additional

1 answer:

3 0

There is not one distance where the cost would be the same. The reason is because the price per mile is the same, but the extra charges are different. This means that no matter when the distance is the same, the cost would always be different.

You might be interested in

he expression 34×12 represents a fraction of a circle that is shaded. Which diagram shows the correct shading?

Tomtit [17]

what is the shading

Step-by-step explanation:

???????

5 0

3 years ago

How to you write 1/25 using exponents

Montano1993 [528]

Answer:

$5^{-2}$

Step-by-step explanation:

Using the rule of exponents

$a^{-m}$ ⇔ $\frac{1}{a^{m} }$ , then

$\frac{1}{25}$ = $\frac{1}{5^{2} }$ = $5^{-2}$

5 0

3 years ago

Read 2 more answers

(1 point) Two six-sided dice are rolled (one red one and one green one). Some possibilities are (Red

anygoal [31]

Complete question:

Two six-sided dice are rolled (one red one and one green one). Some possibilities are (Red= 1, Green=5) or (Red=2, Green=2) etc.

a)How many total possibilities are there?

For the rest of the questions, we will assume that the dice are fair and that all the possibilities in (a) are equally likely.

b)What is the probability that the sum on the two dice comes out to be 11?

c)What is the probability that the sum on the two dice comes out to be 7?

d)What is the probability that the numbers on the two dice are equal?

Answer:

Kindly check explanation

Step-by-step explanation:

Total probabilities :

Total possible outcomes on each dice = 6

Total probabilities = 6² = 36

b)What is the probability that the sum on the two dice comes out to be 11?

Total required outcome = 2

Total possible outcomes = 36

Probability = required outcome / Total possible outcomes

P(sum on both dice = 11) = 2 / 36 = 1/18

c)What is the probability that the sum on the two dice comes out to be 7?

Required outcome = 6

Possible outcomes = 36

P(obtaining a sum of 7) = 6/36 = 1/6

d)What is the probability that the numbers on the two dice are equal

= required outcome / possible outcomes

Required outcome = 1,1 ; 2,2; 3,3; 4,4 ; 5,5 ; 6,6

= 6/36 = 1/6

4 0

3 years ago

What number should go in box A on this number line? 4. 5. A

cestrela7 [59]

Answer:

Step-by-step explanation:

4 0

3 years ago

Solve the following initial-value problem, showing all work, including a clear general solution as well as the particular soluti

Vikki [24]

Answer:

General Solution is $y=x^{3}+cx^{2}$ and the particular solution is $y=x^{3}-\frac{1}{2}x^{2}$

Step-by-step explanation:

$x\frac{\mathrm{dy} }{\mathrm{d} x}=x^{3}+3y\\\\Rearranging \\\\x\frac{\mathrm{dy} }{\mathrm{d} x}-3y=x^{3}\\\\\frac{\mathrm{d} y}{\mathrm{d} x}-\frac{3y}{x}=x^{2}$

This is a linear diffrential equation of type

$\frac{\mathrm{d} y}{\mathrm{d} x}+p(x)y=q(x)$ ..................(i)

here $p(x)=\frac{-2}{x}$

$q(x)=x^{2}$

The solution of equation i is given by

$y\times e^{\int p(x)dx}=\int e^{\int p(x)dx}\times q(x)dx$

we have $e^{\int p(x)dx}=e^{\int \frac{-2}{x}dx}\\\\e^{\int \frac{-2}{x}dx}=e^{-2ln(x)}\\\\=e^{ln(x^{-2})}\\\\=\frac{1}{x^{2} } \\\\\because e^{ln(f(x))}=f(x)]\\\\Thus\\\\e^{\int p(x)dx}=\frac{1}{x^{2}}$

Thus the solution becomes

$\tfrac{y}{x^{2}}=\int \frac{1}{x^{2}}\times x^{2}dx\\\\\tfrac{y}{x^{2}}=\int 1dx\\\\\tfrac{y}{x^{2}}=x+c$ $y=x^{3}+cx^{2$

This is the general solution now to find the particular solution we put value of x=2 for which y=6

we have $6=8+4c$

Thus solving for c we get c = -1/2

Thus particular solution becomes

$y=x^{3}-\frac{1}{2}x^{2}$

5 0

4 years ago