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3 years ago

3x^2 - 4X -3 discreet or continuous

1 answer:

6 0

Discrete And this how A discrete variable is a variable whose value is obtained by counting. A continuous variable is a variable whose value is obtained by measuring. A random variable is a variable whose value is a numerical outcome of a random phenomenon. A discrete random variable X has a countable number of possible values.

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Solve equation by using the quadratic formula.

Sati [7]

Answer: 1: x=3, x=1

2: x= -5

3: There are 2 real solutions.

4: There are 2 real solutions.

5: There are no real solutions.

6. There is 1 real solution.

8. x= -6, x = -2

9. x = -1/6, x=1

10.

Explanation:

1. The quadratic formula is

Substituting our known information we have:

2. Rewriting the quadratic in standard form we have x²+10x-25=0. Substituting this into the quadratic formula gives us:

3. The discriminant is b²-4ac. For this problem, that is 20²-4(-4)(25)=400--400=800. Since this is greater than 0, there are 2 real solutions.

4. The discriminant in this problem is 7²-4(2)(-15)=49--120=49+120=169. This is greater than 0, so there are 2 real solutions.

5. The discriminant in this problem is 1²-4(-2)(-28)=1-224=-223. Since this is less than 0, there are no real solutions.

6. If the discriminant of a quadratic is 0, then by definition there is 1 real solution.

7. Rewriting the quadratic we have 3x²-4x-2=0. Using the quadratic formula we have:

8. Factoring this trinomial we want factors of 12 that sum to 8. 6*2 = 12 and 6+2=8, so those are our factors. This gives us:

(x+6)(x+2)=0

Using the zero product property we know that either x+6=0 or x+2=0. Solving these equations we get x= -6 or x= -2.

9. Factoring this trinomial we want factors of 6(-1)=-6 that sum to -5. (-6)(1)=-6 and -6+1=-5, so this is how we "split up" the x term:

6x²-6x+1x-1=0

We group together the first two and the last two terms:

(6x²-6x)+(1x-1)=0

Factor the GCF out of each group. In the first group, that is 6x:

6x(x-1)+(1x-1)=0

In the second group, the GCF is 1:

6x(x-1)+1(x-1)=0

Both terms have a factor of (x-1), so we can factor it out:

(x-1)(6x+1)=0

Using the zero product property, we know either x-1=0 or 6x+1=0. Solving these equations we get x=1 or x=-1/6.

10. Substituting our information into the quadratic formula we get:

Step-by-step explanation:

5 0

2 years ago

Determine the ordered pair that satisfies the equation, 5x + 2y = 7.

AveGali [126]

Answer:

None of the first 4 do. so I guess the answers E.

Step-by-step explanation:

8 0

3 years ago

Which symbol makes the following TRUE?

olchik [2.2K]

Option A:

6 – 4(3) < 5 – 10

Solution:

Given expression is 6 – 4(3) ____ 5 – 10.

Let us first simplify the expression.

Take left side expression.

⇒ 6 – 4(3)

⇒ 6 – 12

⇒ –6

Take right side expression.

⇒ 5 – 10

⇒ –5

To find which symbol makes the expression true.

In negative the highest number is the lowest in value.

–6 < –5

⇒ 6 – 4(3) < 5 – 10

Option A is the correct answer.

Hence 6 – 4(3) < 5 – 10.

8 0

3 years ago

Francis paid $36.54 for a computer software program after a fifteen percent discount was applied to the original selling price.

tigry1 [53]

Answer:

$42.02

Step-by-step explanation:

15% discount therefore if you want to find the original price it would be 100% plus 15% = 115%

Multiply the percentage (convert to decimal first) with the discounted value:

1.15(36.54) = 42.021

3 0

3 years ago

What set of reflections would carry trapezoid ABCD onto itself? (1 point) Trapezoid ABCD is shown. A is at negative 5, 1. B is a

amid [387]

Answer:

y=x, x axis, y=x, y axis

Step-by-step explanation:

If a point A(x, y) is reflected along the y = x line, the new coordinates would be A'(y, x)

If a point A(x, y) is reflected across the x axis, the new coordinates would be A'(x, -y)

If a point A(x, y) is reflected along the y axis, the new coordinates would be A'(-x, y)

Firstly reflect Trapezoid ABCD along the y = x line to give:

A (-5, 1) ⇒ A'(1, -5). B (-4, 3) ⇒ B'(3, -4). C (-2, 3) ⇒C'(3, -2). D (-1, 1) ⇒ D'(1, -1)

Secondly reflect along the x axis to give:

A'(1, -5)⇒ A"(1,5). B'(3, -4)⇒ B"(3,4), C'(3, -2)⇒C"(3, 2), D'(1, -1) ⇒ D"(1, 1)

Thirdly reflect along the y = x line to give:

A"(1,5) ⇒ A"'(5, 1). B"(3,4) ⇒ B"'(4, 3). C"(3, 2) ⇒ C"'(2, 3). D"(1, 1) ⇒ D"'(1, 1)

Lastly reflect along the y axis to give:

A"'(5, 1) ⇒ A""(-5, 1), B"'(4, 3) ⇒ B""(-4, 3), C"'(2, 3) ⇒ C""(-2, 3), D"'(1, 1) ⇒ D""(-1.1)

6 0

4 years ago