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3 years ago

The graph of an equation with a negative discriminant always has what characteristic

Mathematics

1 answer:

Jet001 [13]3 years ago

5 0

Answer:

See below.

Step-by-step explanation:

The zeroes of the equation will not be real so the graph will not pass through the x-axis.

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Express the fraction in simplest form.

blagie [28]

4/16 in simplest form:

First, we need to find the greatest common factor (GCF) of 4 and 16. Why? We have to find the number that we are going to be dividing the numerator and denominator by to get the simplest form of the fraction. Let's list the factors for 4 and 16:

Factors of 4: 1, 2, 4
Factors of 16: 1, 2, 4, 8, 16

Out of the two sets of factors, which ones are the common ones? The common factors between the two are 1, 2, and 4. Since we are looking for the GREATEST common factor, we have to look at the highest number out of 1, 2, and 4. The GCF is 4, since 4 is higher than 1 and 2.

Second, we now have a number that we are going to divide the numerator and denominator by, which is 4. The numerator is the top number of the fraction (4) and the denominator is the bottom number of the fraction (16).Let's divide now.

$4 \div 4 = 1 \\ 16 \div 4 = 4$

Third, we now have the simplified fraction. (1) is our new numerator, and (4) is our new denominator. This makes the new simplified fraction 1/4. So, your answer is C.

Answer in fraction form: $\fbox {1/4}$
Answer in decimal form: $\fbox {0.25}$

4 0

3 years ago

Read 2 more answers

Please help me out with this

Masteriza [31]

Answer:

C.) x≥ -2/3

Step-by-step explanation:

7 0

3 years ago

Write two division problems that have zero in the quotient. one of the problems should have a remainder and the other should not

kolbaska11 [484]

Yes that is it is 8djudiejeh

8 0

3 years ago

Find each product. Simplify if possible 2/3 x 5/6 x 14=

Ksenya-84 [330]

The product of the probllem is 21

7 0

3 years ago

I need help with pre calculus.

otez555 [7]

Question 1

<h3>Answer: Choice A) -7</h3>

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Explanation:

To compute f(g(2)), we'll first need to find g(2).

Locate 2 in the x row. Then read straight down until you get to the g(x) row. You should find that g(2) = 4

This means f(g(2)) becomes f(4). We then repeat the same process as before. Locate 4 in the x row, read straight down til you get to the f(x) row to find that f(4) = -9

Overall, we can say f(g(2)) = -9

We do the same set of steps to compute g(f(-1)). You should find that f(-1) = -1 and g(f(-1)) = g(-1) = -2

Lastly, we subtract the results we got:

f(g(2)) - g(f(-1)) = -9 - (-2) = -9+2 = -7

==============================================

Question 2

<h3>Answer: Choice B) 0</h3>

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Explanation:

If you were to graph each equation, you should get what you see below.

The red and blue curves do not intersect in any way. We need an intersection point to have a solution. Because there are no intersections, we don't have any solutions. The system is considered inconsistent.

==============================================

Question 3

<h3>Answer: Choice B</h3><h3> $-2x^2+250x-2000$ </h3>

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Explanation:

x = number of scooters made and sold

Each scooter costs $150 to make. If you made x of them, then it costs 150x dollars. Then tack on the fixed cost of $2000 to get the expression $150x+2000$

The cost function is $C(x) = 150x+2000$ . We'll subtract this from the revenue function R(x) to get the profit P(x)

$\text{Profit} = \text{Revenue} - \text{Cost}\\\\P(x) = R(x) - C(x)\\\\P(x) = (400x-2x^2) - (150x+2000)\\\\P(x) = 400x-2x^2-150x-2000\\\\P(x) = -2x^2+250x-2000\\\\$

==============================================

Question 4

<h3>Answer: Choice B</h3>

$\left\{x\in \mathbb{R} \big| \ x \ne -2, x \ne 1\right\}$

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Explanation:

$g(x) = \frac{1}{x+1}\\\\g(f(x)) = \frac{1}{f(x)+1}\\\\g(f(x)) = \frac{1}{x^2+x-3+1}\\\\g(f(x)) = \frac{1}{x^2+x-2}\\\\g(f(x)) = \frac{1}{(x+2)(x-1)}\\\\(g\circ f)(x) = \frac{1}{(x+2)(x-1)}\\\\$

I started with the outer function g(x) and replaced each x with f(x) in the second step. Then I plugged in f(x) = x^2+x-3 and simplified. Factoring the denominator is helpful to determine where the denominator becomes zero. Specifically, it happens when x = -2 and x = 1. These x values must be kicked out of the domain to avoid a division by zero error.

So we start off with the real number line, and poke holes at x = -2 and x = 1.

The notation $x \in \mathbb{R}$ means x is a real number. Then we tack on $x \ne -2, x \ne 1$

Overall, the domain is $\left\{x\in \mathbb{R} \big| \ x \ne -2, x \ne 1\right\}$

Let's see what happens if we tried plugging x = -1 into this composite function:

$(g \circ f)(x) = \frac{1}{(x+2)(x-1)}\\\\(g \circ f)(-1) = \frac{1}{(-1+2)(-1-1)}\\\\(g \circ f)(-1) = \frac{1}{(1)(-2)}\\\\(g \circ f)(-1) = \frac{1}{-2}\\\\(g \circ f)(-1) = -0.5\\\\$

We don't get a division by zero error, so x = -1 is allowed. The same can be said about any other x value as long as it's not -2 and not 1 either.

4 0

2 years ago