I need help!!!!! Help Needed!!!!!!!!!!!

Makovka662 [10]

Answer:

y = 2x² - 5x + 7

Step-by-step explanation:

For this problem, imagine the standard form, which is:

f(x) = ax² + bx + c = y

Using the table, and adding that y at the end, we can plug in and write out the following equations. (Writing it out is important):

f(-1) = a(-1)² + b(-1) + c = 14

f(0) = a(0)² + b(0) + c = 7

f(1) = a(1)² + b(1) + c = 4

f(2) = a(2)² + b(2) + c = 5

Now if this doesn't look familiar, it's actually a systems of equations using the a, b, and c elements as your three variables! If you simplify the equations:

f(-1) = a - b + c = 14

f(0) = c = 7

f(1) = a + b + c = 4

f(2) = 4a + 2b + c = 5

Something unique just happened. We have already defined what ' c ' is!

c = 7

Setting that aside, if you remove the f(x) portion of the equations, you're left with:

a - b + c = 14

a + b + c = 4

4a + 2b + c = 5

Using the two upper equations, if we add them together (you can do that as it doesn't change the values of the variables) you get:

2a + 2c = 18

Note: the ' b ' variables cancelled out in the addition [ b + (-b) ]

If you further simplify the equation:

a + c = 9

Awesome. Now we already know that c = 7, so if you plug that into the equation:

a + 7 = 9

Solve for a. So then a = 2

Now that we know the following:

a = 2

c = 7

We can then use the equation:

a + b + c = 4

And solve for b!

2 + b + 7 = 4

Simplify.

b + 9 = 4

Simplify.

b = -5

Now at this point, since you know what a, b and c are, you can write the equation!

f(x) = 2x² - 5x + 7

You can confirm your work by putting any of the x values in the table through!

4 0

3 years ago

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Third grade 1 - 4/7 =?

Hitman42 [59]

Answer:

3/7

or

0.4285714286

4 0

3 years ago

Read 2 more answers

If 8 identical blackboards are to be divided among 4 schools,how many divisions are possible? How many, if each school mustrecei

MAXImum [283]

Answer:

There are 165 ways to distribute the blackboards between the schools. If at least 1 blackboard goes to each school, then we only have 35 ways.

Step-by-step explanation:

Essentially, this is a problem of balls and sticks. The 8 identical blackboards can be represented as 8 balls, and you assign them to each school by using 3 sticks. Basically each school receives an amount of blackboards equivalent to the amount of balls between 2 sticks: The first school gets all the balls before the first stick, the second school gets all the balls between stick 1 and stick 2, the third school gets the balls between sticks 2 and 3 and the last school gets all remaining balls.

The problem reduces to take 11 consecutive spots which we will use to localize the balls and the sticks and select 3 places to put the sticks. The amount of ways to do this is ${11 \choose 3} = 165 .$ As a result, we have 165 ways to distribute the blackboards.

If each school needs at least 1 blackboard you can give 1 blackbooard to each of them first and distribute the remaining 4 the same way we did before. This time there will be 4 balls and 3 sticks, so we have to put 3 sticks in 7 spaces (if a school takes what it is between 2 sticks that doesnt have balls between, then that school only gets the first blackboard we assigned to it previously). The amount of ways to localize the sticks is ${7 \choose 3} = 35$ . Thus, there are only 35 ways to distribute the blackboards in this case.