Which is the absolute value function corresponding to the graph with these characteristics?

2 answers:

8 0

The parent function for an absolute value function is:

$y=a|x-h|+k$

If a is positive, the graph will open upward, and if a is negative, it will open downward. Given that the graph opens downward, we know a must be negative.

Next, h represents the horizontal shift of the graph. Given that the vertex is at (-2,k), we know that h = 2.

The k value does not matter because in the given characteristics we are given a general k value that will work with any graph.

Finally, we need to find which graph passes through (-3, -1). To meet the requirements discussed above, we can narrow the answer to C or D. Now just plug in -3 for x in both functions, and whichever one equals -1 at x=-3 is our answer:

C) -4|-3 + 2| + 3 = -4(1) + 3 = -1

D) -4|-3 + 2| - 3 = -4(1) - 3 = -7

The answer is C.

Anton [14]3 years ago

3 0

The correct answer would be C

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Rina is making a nut mixture to sell at the local farmers' market. She mixes 6 pounds of pecans with a nut mixture that is 70% p

Marat540 [252]

So the best way to do these is concentration1 (%) × volume1 = concentration2 × volume2
Or C1V1 + C2V2 = C3V3, where C1 = 100% (bc ALL pecans), V1 = 6 lbs, C2 = 70%, C3 = 82%:
100%×6 + 70%×v2 = 82%×(6+v2)
100%=1.00, 70%=.7, 82%=.82
note: if none is poured out then v3 = v1+v2
6 + .7v2 = .82 (6+v2)
6 + .7v2 = 4.92 + .82v2
6 + .7v2 -.7v2 = 4.92 + .82v2 -.7v2
6 = 4.92 + .12v2
6-4.92 = 4.92-4.92 + .12v2
1.08 = .12v2
.12v2/.12 = 1.08/.12
v2 = 9 lbs
that's only v2!!!

For the final poundage, we need v3:

v3 = 6 + v2 = 6 + 9 = 15 lbs

7 0

3 years ago

Two dance teams, the Roses and the Tulips, are perfectly matched so that their chances of winning are both set at 50%. What is t

shtirl [24]

Answer:

Step-by-step explanation:

Step 1

The fist step is to define the probabilities.

The events are independent from each other. Each team wins with probability $\frac{1}{2}$ and looses with probability $\frac{1}{2}.$ Let $P(W_R)$ be probability that Roses win and $P(L_R)$ be the probability that Roses loose.

Step 2

The second step is to calculate the probabilities by multiplying the probabilities with the first 3 terms in the product being the probability of a win and the last term being the probability of a loss.

The calculation is shown below,

$P(W_R)\times P(W_R)\times P(W_R)\times P(L_R)=\frac{1}{2} \times \frac{1}{2} \times\frac{1}{2} \times\frac{1}{2} =\frac{1}{16}$