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inna [77]
3 years ago
6

Why do teachers give out so much homework? I can understand why colledge and hugh school get it, but why the younger ones? They'

re still delevoping. Why? (This doesn't have to do with any subjects; it's not math)...
Mathematics
1 answer:
seraphim [82]3 years ago
8 0
Most teachers say that they do it to make sure the kids dont't forget what ws tought that day or to review
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Mia paid $21.00 to rent 12 games throughout the month. The store charges $1.25 for new games and $2.75 for older games. How many
zepelin [54]

Answer:

8

Step-by-step explanation:

1.25x8+2.75x4

7 0
3 years ago
Someone help me with this please:)
Tom [10]

Answer:

480 ft2

Step-by-step explanation:

You multipy 4 by 2 by 2 by 10 by 3 and you get 480.

Hope this helps :)

3 0
3 years ago
Read 2 more answers
I need the answers plz
garik1379 [7]

Answer:

These problems are an example of equations with two unknowns. The way these equations are solved is that we write these equations one under the another.

If both equations have, such is the case here, same parts, we can simply cancel the same parts out and subtract the rest of equatuons. That way, we are left with only one unknown (the other one was eliminated), which makes it easy to solve.

After we have found the value of an unknown, we just plug it back into any of the starting equations and solve for the second unknown.

2. Adult ticket costs $12 and child ticket costs $14.

3. Adult ticket costs $10 and child ticket costs $5.

4. One daylily costs $9 and one bush of ornamental grass costs $2.

5. A van can carry 15 and a bus can carry 56 students.

Step-by-step explanation:

2. If we mark the price of one adult ticket with x and the price of one child ticket with y, we get that:

- first day: 7x + 12y = $252

- second day: 7x + 10y = $224

Now, we can make a system:

7x + 12y = 252

7x + 10y = 224

We can now subtract these two equations and 7x will cancel out, so we get:

12y - 10y = 252 - 224

2y = 28

y = 14

Now, we can plug the value of y into any of the two equations:

7x + 10y = 224

7x + 140 = 224

7x = 84

x = 12

3. Similarly, if we mark the price of one adult ticket with x and the price of one child tickey with y, we'll get a system:

x + 12y = 70

x + 9y = 55

Again, if we subtract these two, x will cancel out, so we have:

12y - 9y = 70 - 55

3y = 15

y = 5

Now, we plug the value of y into any of the two equations, and we get:

x + 9y = 55

x + 45 = 55

x = 10

4. Using the same principle, we can mark the price of one daylily with x and the price of one bunch of ornamental grass with y, we'll get a system:

12x + 11y = 130

12x + 12y = 132

Again, we subtract so that 12x cancel out and we get:

11y - 12y = 130 - 132

-y = -2

If we get minuses on both sides, we can simply multiply both sides with -1 and we get:

y = 2

Again, we plug y:

12x + 12y = 132

12x + 24 = 132

12x = 108

x = 9

5. If we mark number of students in a van with x and the number of students in a bus with y, we get a system:

2x + 12y = 702

2x + y = 86

As you've probably already noticed the pattern, we subtract equations and cancel 2x out to get:

12y - y = 702 - 86

11y = 616

y = 56

Once again, we plug the value of y into any equation:

2x + y = 86

2x + 56 = 86

2x = 30

x = 15

4 0
3 years ago
Which graph represents the piecewise-defined function?
chubhunter [2.5K]

Answer: The correct option is 1.

Explanation:

The given piecewise function is,

y=\begin{cases}-\frac{4}{5}x-3 & \text{ if } x

It means if x<0, then

f(x)=-\frac{4}{5}x-3

If x\geq2, then

f(x)=3x-10

Since the f(x) is defined for x<0 and x\geq2, therefore the function f(x) is not defined for 0\leq x.

From the graph 2, 3 and 4 we can easily noticed that for each value of x there exist a unique value of y, therefore the function is defined for all values of x, which is not true according to the given piecewise function.

Only in figure the value of y not exist when x lies between 0 to 2, including 0. It means the function is not defined for 0\leq x, hence the first option is correct.

6 0
4 years ago
In your own words explain why a quadratic<br> equation can't have one imaginary solution.
Masja [62]
A quadratic equation cannot have one imaginary solution because of the discriminant enclosed in a radical. The discriminant, √(b² - 4ac), determines the nature of the roots and it can only be either 2 real roots, 1 real solution or 2 imaginary roots. Hope this answers the question.
7 0
3 years ago
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