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Brrunno [24]
3 years ago
11

Please help me with this!!!

Mathematics
1 answer:
Daniel [21]3 years ago
8 0
You have it correct already
You might be interested in
10 POINTS!!! FULL ANSWER IN STEP BY STEP FORMAT!!
stich3 [128]
A) The signs of the first derivative (g') tell you the graph increases as you go left from x=4 and as you go right from x=-2. Since g(4) < g(-2), one absolute extreme is (4, g(4)) = (4, 1).

The sign of the first derivative changes at x=0, at which point the slope is undefined (the curve is vertical). The curve approaches +∞ at x=0 both from the left and from the right, so the other absolute extreme is (0, +∞).

b) The second derivative (g'') changes sign at x=2, so there is a point of inflection there.

c) There is a vertical asymptote at x=0 and a flat spot at x=2. The curve goes through the points (-2, 5) and (4, 1), is increasing to the left of x=0 and non-increasing to the right of x=0. The curve is concave upward on [-2, 0) and (0, 2) and concave downward on (2, 4]. A possible graph is shown, along with the first and second derivatives.

8 0
3 years ago
Simplify \sqrt{0.81}
olchik [2.2K]

Answer:

Square root of .81 = .9

7 0
4 years ago
I need help with pre calculus.
otez555 [7]

Question 1

<h3>Answer:  Choice A) -7</h3>

-----------------

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>

-----------------

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>

-----------------

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\}

-----------------

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
The absolute value function g(x) = |x + 7| − 4 is translated 5 units right and 2 units up to become g′(x). The quadratic functio
madreJ [45]
I think its C, if I'm being quite honest its kind of confusing for me too. Sorry
4 0
3 years ago
Read 2 more answers
Solve then check solution
klio [65]

Answer:

1. z = 128

2. x = 4.2

3. c = 10

4. w = 100

5. a = 95.2

Step-by-step explanation:

1. Solve for z:

z/16 = 8

Multiply both sides of z/16 = 8 by 16:

(16 z)/16 = 16×8

(16 z)/16 = 16/16×z = z:

z = 16×8

16×8 = 128:

Answer:  z = 128

_____________________________________________________

2. Solve for x:

3.5 x = 14.7

Divide both sides of 3.5 x = 14.7 by 3.5:

(3.5 x)/3.5 = 14.7/3.5

3.5/3.5 = 1:

x = 14.7/3.5

14.7/3.5 = 4.2:

Answer:  x = 4.2

____________________________________________

3. Solve for c:

32 = 3.2 c

32 = 3.2 c is equivalent to 3.2 c = 32:

3.2 c = 32

Divide both sides of 3.2 c = 32 by 3.2:

(3.2 c)/3.2 = 32/3.2

3.2/3.2 = 1:

c = 32/3.2

32/3.2 = 10:

Answer:  c = 10

__________________________________________________

4. Solve for w:

(2 w)/5 = 40

Multiply both sides of (2 w)/5 = 40 by 5/2:

(5×2 w)/(2×5) = 5/2×40

5/2×2/5 = (5×2)/(2×5):

(5×2)/(2×5) w = 5/2×40

5/2×40 = (5×40)/2:

(5×2 w)/(2×5) = (5×40)/2

(5×2 w)/(2×5) = (2×5)/(2×5)×w = w:

w = (5×40)/2

2 | 2 | 0

| 4 | 0

- | 4 |  

| | 0

| - | 0

| | 0:

w = 5×20

5×20 = 100:

Answer: w = 100

___________________________________________________

5. Solve for a:

a/14 = 6.8

Multiply both sides of a/14 = 6.8 by 14:

(14 a)/14 = 14×6.8

(14 a)/14 = 14/14×a = a:

a = 14×6.8

14×6.8 = 95.2:

Answer:  a = 95.2

6 0
3 years ago
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