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3 years ago

11

An internet service provider charges $1 per GB plus a $12 flat fee. If: • $$ represents the number of GB used • $$ represents th

e total cost What equation models this situation

1 answer:

nordsb [41]3 years ago

7 0

Answer:

Step-by-step explanation:

The question isn’t displaying properly, so I’ll use My own formatting.

y is the total cost in dollars, x is the number of gigabytes used

y = x + 12

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Easy question, please help

stepladder [879]

Step-by-step explanation:

All you need to do is divide cost by mile since they already factored in the additional $3 for you.

1. 8/5 = $1.60

2. 11/8 = $1.375 = $1.38

3. 23/20 = $1.15

6 0

2 years ago

A^2 + b^2 + c^2 = 2(a − b − c) − 3. (1) Calculate the value of 2a − 3b + 4c.

Verdich [7]

Answer:

$2a - 3b + 4c = 1$

Step-by-step explanation:

Given

$a^2 + b^2 + c^2 = 2(a - b - c) - 3$

Required

Determine $2a - 3b + 4c$

$a^2 + b^2 + c^2 = 2(a - b - c) - 3$

Open bracket

$a^2 + b^2 + c^2 = 2a - 2b - 2c - 3$

Equate the equation to 0

$a^2 + b^2 + c^2 - 2a + 2b + 2c + 3 = 0$

Express 3 as 1 + 1 + 1

$a^2 + b^2 + c^2 - 2a + 2b + 2c + 1 + 1 + 1 = 0$

Collect like terms

$a^2 - 2a + 1 + b^2 + 2b + 1 + c^2 + 2c + 1 = 0$

Group each terms

$(a^2 - 2a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

Factorize (starting with the first bracket)

$(a^2 - a -a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$(a(a - 1) -1(a - 1)) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1) (a - 1)) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b^2 + b+b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b(b + 1)+1(b + 1)) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)(b + 1)) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c^2 + c+c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c(c + 1)+1(c + 1)) = 0$

$((a - 1)^2) + ((b + 1)^2) + ((c + 1)(c + 1)) = 0$

$((a - 1)^2) + ((b + 1)^2) + ((c + 1)^2) = 0$

Express 0 as 0 + 0 + 0

$(a - 1)^2 + (b + 1)^2 + (c + 1)^2 = 0 + 0+ 0$

By comparison

$(a - 1)^2 = 0$

$(b + 1)^2 = 0$

$(c + 1)^2 = 0$

Solving for $(a - 1)^2 = 0$

Take square root of both sides

$a - 1 = 0$

Add 1 to both sides

$a - 1 + 1 = 0 + 1$

$a = 1$

Solving for $(b + 1)^2 = 0$

Take square root of both sides

$b + 1 = 0$

Subtract 1 from both sides

$b + 1 - 1 = 0 - 1$

$b = -1$

Solving for $(c + 1)^2 = 0$

Take square root of both sides

$c + 1 = 0$

Subtract 1 from both sides

$c + 1 - 1 = 0 - 1$

$c = -1$

Substitute the values of a, b and c in $2a - 3b + 4c$

$2a - 3b + 4c = 2(1) - 3(-1) + 4(-1)$

$2a - 3b + 4c = 2 +3 -4$

$2a - 3b + 4c = 1$

7 0

4 years ago

-10-(-6) solve the integers

Ghella [55]

Answer:

It would be -4

Step-by-step explanation:

a double negative becomes a + sign

So, -10 + 6=-4

4 0

3 years ago

Read 2 more answers

a test has 20 multiple-choice questions with 5 choices each, followed by 35 true/false questions. if a student guesses on each q

Nuetrik [128]

Answer

There are 170 ways the student can answer the test.

Explanation

If there are 20 multiple-choice questions with 5 choices each, the student has 100 choices. The first question has 5 choices to pick from. The second has 5 as well. So does the third. Hopefully now you realize that you have to multiply the number of choices by the number of questions.

The same thing goes with the true/false questions. There are 2 choices for each true/false question, and there are 35 of those. 35×2 is 70. There are 70 ways to answer on the true/false questions.

Now combine the number of choices on the first part and the second part; 100+70 is 170.

3 0

3 years ago

Is one third rational or irrational

Zinaida [17]

One third is a rational number....since it can be written in p/q form(1/3)

5 0

3 years ago

Read 2 more answers