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nataly862011 [7]
3 years ago
15

PLEASE ANSWER! DOING ANOTHER QUICK TEST! REAL QUICK! MUST BE CORRECT! POINTS WILL BE GIVEN!

Mathematics
1 answer:
Alexxandr [17]3 years ago
6 0
-6 * (-20)

  -6
-20
-------
120 (Two negatives = a positive!)
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Answer: C

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Make a table for the following equations<br> y = 2x + 4
nekit [7.7K]

In order to make a table, we sample some x values (whichever we want), and we compute the expression for those value. Each x value will yield a unique y value.

If you need this table to graph the function, you'll only need two points, since this is a line, and having two points you just need to connect them.

Here are some samples, feel free to make more if you need to:

f(x)=2x+4\implies f(0)=2\cdot 0 + 4 = 4

f(x)=2x+4\implies f(1)=2\cdot 1 + 4 = 6

f(x)=2x+4\implies f(2)=2\cdot 2 + 4 = 8

f(x)=2x+4\implies f(3)=2\cdot 3 + 4 = 10

f(x)=2x+4\implies f(4)=2\cdot 4 + 4 = 12

So, we have the following table

\begin{array}{c|c}0&4\\1&6\\2&8\\3&10\\4&12\end{array}

4 0
3 years ago
Solve for q. -5(3-q) +4=5q-11
soldier1979 [14.2K]

Perform the indicated multiplication first:  -15 + 5q + 4 = 5q - 11

Note that 5q appears on both sides of this equation.  Cancelling, we get :

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4 0
2 years ago
Calculate the discriminant to determine the number solutions. y = x ^2 + 3x - 10
Nataly_w [17]

1. The first step is to find the discriminant itself. Now, the discriminant of a quadratic equation in the form y = ax^2 + bx + c is given by:

Δ = b^2 - 4ac

Our equation is y = x^2 + 3x - 10. Thus, if we compare this with the general quadratic equation I outlined in the first line, we would find that a = 1, b = 3 and c = -10. It is easy to see this if we put the two equations right on top of one another:

y = ax^2 + bx + c

y = (1)x^2 + 3x - 10

Now that we know that a = 1, b = 3 and c = -10, we can substitute this into the formula for the discriminant we defined before:

Δ = b^2 - 4ac

Δ = (3)^2 - 4(1)(-10) (Substitute a = 1, b = 3 and c = -10)

Δ = 9 + 40 (-4*(-10) = 40)

Δ = 49 (Evaluate 9 + 40 = 49)

Thus, the discriminant is 49.

2. The question itself asks for the number and nature of the solutions so I will break down each of these in relation to the discriminant below, starting with how to figure out the number of solutions:

• There are no solutions if the discriminant is less than 0 (ie. it is negative).

If you are aware of the quadratic formula (x = (-b ± √(b^2 - 4ac) ) / 2a), then this will make sense since we are unable to evaluate √(b^2 - 4ac) if the discriminant is negative (since we cannot take the square root of a negative number) - this would mean that the quadratic equation has no solutions.

• There is one solution if the discriminant equals 0.

If you are again aware of the quadratic formula then this also makes sense since if √(b^2 - 4ac) = 0, then x = -b ± 0 / 2a = -b / 2a, which would result in only one solution for x.

• There are two solutions if the discriminant is more than 0 (ie. it is positive).

Again, you may apply this to the quadratic formula where if b^2 - 4ac is positive, there will be two distinct solutions for x:

-b + √(b^2 - 4ac) / 2a

-b - √(b^2 - 4ac) / 2a

Our discriminant is equal to 49; since this is more than 0, we know that we will have two solutions.

Now, given that a, b and c in y = ax^2 + bx + c are rational numbers, let us look at how to figure out the number and nature of the solutions:

• There are two rational solutions if the discriminant is more than 0 and is a perfect square (a perfect square is given by an integer squared, eg. 4, 9, 16, 25 are perfect squares given by 2^2, 3^2, 4^2, 5^2).

• There are two irrational solutions if the discriminant is more than 0 but is not a perfect square.

49 = 7^2, and is therefor a perfect square. Thus, the quadratic equation has two rational solutions (third answer).

~ To recap:

1. Finding the number of solutions.

If:

• Δ < 0: no solutions

• Δ = 0: one solution

• Δ > 0 = two solutions

2. Finding the number and nature of solutions.

Given that a, b and c are rational numbers for y = ax^2 + bx + c, then if:

• Δ < 0: no solutions

• Δ = 0: one rational solution

• Δ > 0 and is a perfect square: two rational solutions

• Δ > 0 and is not a perfect square: two irrational solutions

6 0
3 years ago
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