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Rufina [12.5K]
2 years ago
12

HELP PLZ IM CRYING I DONT WANT TO FAIL MY TEST PLZ HELP!!!!!!!!!

Mathematics
1 answer:
ivann1987 [24]2 years ago
4 0

Answer:

As the x values go to positive infinity, the functions values go to positive infinity

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HELP PLEASE!!!!!!!!!!!!!!!!!!!!!
devlian [24]
I think it’s the third one and the last two. I’m sorry if it’s wrong
7 0
3 years ago
What is 12 times 12 plus 1
sammy [17]

Answer:

145

Step-by-step explanation:

just use a calculator

3 0
3 years ago
Read 2 more answers
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
tickets for a football match are sold at $30 for adults and $15 for children a company bought 28 tickets if x of these tickets w
anzhelika [568]
First, we must let:
   x = number of tickets intended for adults
   y = number of tickets intended for children.

a. Write in terms of x the number of tickets for children
     Solution:
           x + y = 28   ⇔    y = 28 - x   (equation 1)
       To answer in terms of x:
           no. of tickets for tickets for children = 28 - x

b. the amount spent on tickets for adults
    Solution:  $30 is the cost of ticket per adult and there are x number of tickets intended for adults.
           Therefore, 
            amount spent on ticket for adults = 30x

c. the amount spent on the tickets.
     Solution:
       $ 15  = cost of ticket per child
       $ 30 = cost of ticket per adult

      total amount spent on tickets = 30x + 15y    ⇒   (equation2)
  substitute equation 1 to equation 2.
  (equation 1)   y = 28 - x
  (equation 2)   total amount spent on tickets = 30x + 15y
                        total amount spent on tickets = 30x + 15(28-x)
                        total amount spent on tickets = 30x + 420 - 15x
                        total amount spent on tickets = 15x + 420
4 0
3 years ago
how many circles can be drawn passing through three non collinear points and how many circles can be drawn passing through three
MrRissso [65]

Answer:

Only one circle can be drawn through three non nonlinear points, No circle can be drawn through three nonlinear points

Step-by-step explanation:

A circle has a curve. As long as the three non nonlinear points both have the same amount of distance from the center of the circle it can be a circle drawn. No circle can be drawn through three points on the line.

7 0
2 years ago
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