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3 years ago

One quadratic function has the formula h(x) = -x 2 + 4x - 2. Another quadratic function, g(x), has the graph shown below

Mathematics

2 answers:

jeyben [28]3 years ago

7 0

We are comparing maxima. From the graph we know that the max of one graph is +2 at x = -2. What about the other graph? Need to find the vertex to find the max.

Complete the square of <span>h(x) = -x^2 + 4x - 2:

</span>h(x) = -x^2 + 4x - 2 = -(x^2 - 4x) -2
= -(x^2 - 4x + 4 - 4) - 2
=-(x^2 - 4x + 4) -2+4
= -(x-2)^2 + 2 The equation describing this parabola is y=-(x-2)^2 + 2, from which we know that the maximum value is 2, reached when x = 2.

The 2 graphs have the same max, one at x = -2 and one at x = + 2.

ale4655 [162]3 years ago

7 0

Answer:

Functions g and h have the same maximum of 2.

Step-by-step explanation:

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hram777 [196]

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4 years ago

There are 52 cards in a deck, and 13 of them are hearts. Consider the following two scenarios:

Klio2033 [76]

Answer:

The two scenarios differ.

P(Scenario A) = 0.107

P(Scenario B) = 0.265

Step-by-step explanation:

The two scenarios are different because for the second one you are alredy assuming that the first 3 cards are not hearths, and for that reason, B is more likely to happen that A.

For B to happen, you need to notice that since you remove 3 cards from your deck that are not hearts, then your deck has only 49 cards, and 13 of them are hearts. The probability for a heart to show up is, as a result 13/49 = 0.265 because you have 13 favourable cases from 49 possible.

For A, you need the first card to be anything but a heart. Since 13 cards of the deck are herts, 39 are not, and the probability of that hapening is 39/52 = 3/4. After you remove your first card, the probability of the second one not being heart is 38/51, and the probability for the third one is 37/50 (you are removing one favourable case and one case for the total of cases each time). The probability for the fourth card being a heart assuming that the first three are not was calculated before and it gives us a result of 13/49.

Multiplying everything, we obtain that

P(A) = 3/4*38/51*37/50*13/49 = 0.107.

8 0

3 years ago

Help me with precal pleaseee

yuradex [85]

Answer: Choice C

$\left[0 , \frac{\pi}{2}\right) \ \ \text{ and } \ \ \left(\frac{\pi}{2}, \pi\right]$

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Explanation:

Let's look at the function y = sec(x) first, which is the secant function.

Recall that secant is 1 over cosine, so sec(x) = 1/cos(x)

We can't divide by zero, so cos(x) = 0 can't be allowed. If x = pi/2, then cos(pi/2) = 0 will happen. So we must exclude pi/2 from the domain of sec(x).

If we look at the interval from 0 to pi, then the domain of sec(x) is $0 \le x < \frac{\pi}{2} \ \ \text{ and } \ \ \frac{\pi}{2} < x \le \pi$

we can condense that into the interval notation $\left[0 , \frac{\pi}{2}\right) \ \ \text{ and } \ \ \left(\frac{\pi}{2}, \pi\right]$

Note the use of curved parenthesis to exclude the endpoint; while the square bracket includes the endpoint.

So effectively we just poked at hole at x = pi/2 to kick that out of the domain. I'm only focusing on the interval from 0 to pi so that secant is one to one on this interval. That way we can apply the inverse. When we apply the inverse, the domain and range swap places. So the range of arcsecant, or $\sec^{-1}(x)$ is going to also be $\left[0 , \frac{\pi}{2}\right) \ \ \text{ and } \ \ \left(\frac{\pi}{2}, \pi\right]$

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3 years ago

I’ve looked at this for a while, can someone clear up the answers for me? I can’t figure it out.

navik [9.2K]

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3 years ago

If v = X + 4t, v is the velocity and X is the displacement, how to find X in terms of t?

kobusy [5.1K]

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2 years ago