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Tems11 [23]
4 years ago
13

John rents a car for $85 a week plus $0.26 per mile. How far can he travel on a budget of $215?

Mathematics
1 answer:
Gnom [1K]4 years ago
4 0
John can drive 500 miles. IF he drives all the miles in one week.
215-85=130. 130^0.26 =500
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7/5 + 2/3 - 1 = ayuda porfa
brilliants [131]

Answer:

1\frac{1}{15}

Step-by-step explanation:

\frac{7}{5}+\frac{2}{3}-1\\

Convertir elemento a fracción: 1 = 1/1

=-\frac{1}{1}+\frac{7}{5}+\frac{2}{3}

Mínimo común múltiplo de 1,5,3 = 15

Ajustar fracciones según el mcm

=-\frac{15}{15}+\frac{21}{15}+\frac{10}{15}

Dado que los denominadores son iguales, combine las fracciones:

\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}\\\\= \frac{-15+21+10}{15}\\\\

Sumar Restar los números: -15 + 21 + 10 = 16

=\frac{16}{15}\\\\16\div 15 = 1 \: remainder 1\\\\=1\frac{1}{15}

4 0
3 years ago
Caitlin bought 6 cookies for a total of $10. At this rate, what is the cost of 9 cookies?
ozzi

Answer:

15$

Step-by-step explanation:

cost of 1 cookie = 10/6

cost of 9 cookies = 9* 10/6 = $15

7 0
4 years ago
Read 2 more answers
Let l(x) be the statement "x has an Internet connection" and C(x, y) be the statement "x and y have chatted over the Internet,"
Bas_tet [7]

The question is:

Let l(x) be the statement "x has an Internet connection" and C(x, y) be the statement "x and y have chatted over the Internet," where the domain for the variables x and y consists of all students in your class. Use quantifiers to express each of these statements.

a. Jerry does not have an Internet connection.

b. Rachel has not chatted over the Internet with Chelsea.

c. Jan and Sharon have never chatted over the Internet.

d. No one in the class has chatted with Bob.

e. Sanjay has chatted with everyone except Joseph.

f. Someone in your class does not have an Internet connection.

g. Not everyone in your class has an Internet connection.

h. Exactly one student in your class has an internet connection.

i. Everyone except one student in your class has an Internet connection.

j. Everyone in your class with an Internet connection has chatted over the Internet with at least one other student in your class.

k. Someone in your class has an Internet connection but has not chatted with anyone else in your class.

l. There are two students in your class who have not chatted with each other over the Internet.

m. There is a student in your class who has chatted with everyone in your class over the Internet.

n. There are at least two students in your class who have not chatted with the same person in your class.

o. There are two students in the class who between them have chatted with everyone else in the class.

Step-by-step explanation:

Note that: Because the arguments are variables, we write

Ix = I(x)

Cxy = C(x,y)

a. Jerry does not have an Internet connection.

~I(Jerry).

b. Rachel has not chatted over the internet with Chelsea.

~C(Rachel,Chelsea).

c. Jan and Sharon have never chatted over the internet.

~C(Jan,Sharon).

d. No one in the class has chatted with Bob.

~∃x C(x, Bob).

e. Sanjay has chatted with everyone except Joseph.

∀y (y ≠ Joseph → C(Sanjay, y)).

f. Someone in your class does not have an internet connection.

∃x ~Ix.

g. Not everyone in your class has an internet connection.

~∀x Ix.

h. Exactly one student in your class has an internet connection.

∃x (Ix ∧ ∀y (Iy → y = x)).

i. Everyone except one student in your class has an internet connection. This means exactly one student does not have a connection.

∃x (~Ix∧ ∀y (Iy → y = x)).

j. Everyone in your class with an internet connection has chatted over the internet with at least one other student in your class.

∀x (Ix → ∃y (Cxy ∧ x ≠ y)).

Assuming nobody can chat with him or herself, then

∀x (Ix → ∃y (Cxy))

k. Someone in your class has an internet connection but has not chatted with anyone else in your class.

∃x (Ix ∧ ∀y ~Cxy).

l. There are two students in your class who have not chatted with each other over the internet.

∃x ∃y (x ≠ y∧~Cxy).

m. There is a student in your class who has chatted with everyone in your class over the internet.

∃x ∀y Cxy.

n. There are at least two students in your class who have not chatted with the same person in your class. ∃x ∃y (x ≠ y ∧ ∃z (~Cxz ∧ ~Cyz)).

o. There are two students in the class who between them have chatted with everyone else in the class.

∃x ∃y (x ≠ y ∧ ∀z (Cxz ∨ Cyz)).

5 0
4 years ago
Find the perimeter of the parallelogram shown below
erastovalidia [21]

Answer:

55+55+52+52=214

I HOPE IT WILL HELP YOU. YOU ARE WELCOMMMMMMME.

5 0
3 years ago
Solve the system, or show that it has no solution. (If there is no solution, enter NO SOLUTION. If there are an infinite number
Tanzania [10]

Given:

The equation is

\begin{gathered} 6x+4y=16 \\  \\ 9x+6y=24 \end{gathered}

Find-:

The solution of the equation

Explanation-:

The first equation is

\begin{gathered} 6x+4y=16 \\  \\ 2(3x+2y)=16 \\  \\ 3x+2y=\frac{16}{2} \\  \\ 3x+2y=8 \end{gathered}

The second equation is

\begin{gathered} 9x+6y=24 \\  \\ 3(3x+2y)=24 \\  \\ 3x+2y=\frac{24}{3} \\  \\ 3x+2y=8 \\  \end{gathered}

Both equations are the same.

For slove, the two unknown variables need two equations.

So,

There is NO SOLUTION

8 0
1 year ago
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