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cupoosta [38]
3 years ago
9

Find the limit of the function algebraically. x→−3 (x^2-9/x^3+3)

Mathematics
1 answer:
Naya [18.7K]3 years ago
3 0

Answer:

  d)  0

Step-by-step explanation:

The function is defined for x=-3, so it can simply be evaluated there.

\displaystyle\lim_{x \to -3}\left(\frac{x^2-9}{x^3+3}\right)=\frac{(-3)^2-9}{(-3)^3+3}=\frac{9-9}{-27+3}=0

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What is the range of the function graphed below?
Nadusha1986 [10]
Its the second option!

The range expresses the constraints of y-values of a function. There is a solid point at positive three which means that it is included. The open point at -3 means it’s not included. This makes the range -3
3 0
2 years ago
An analysis of sales records for the last 120 weeks gives the following results. Assuming that these past data are a reliable gu
Mila [183]

Answer:

(a) 0.5333

(b) 0.6583

(c) 0.5583

(d) 0.7083

(e) 0.6167

Step-by-step explanation:

Denote the events as follows:

<em>A</em> = a competitor will advertise

<em>NA </em>= a competitor will not advertise

<em>L </em>= Low sales will be achieved

<em>M </em>= Medium sales will be achieved

<em>H </em>= High sales will be achieved

The data provided is of the form:

          Low (L)    Medium (M)    High (H)    Total    

A            32                   14                 18               64

NA          21                   12                 23              56

Total      53                  26                 41              120

The probability of an event <em>E</em> is:

P(E)=\frac{n(E)}{N}

n (E) = favorable outcomes of event <em>E</em>

N = Total number of outcomes

(a)

Compute the probability that next week the competitor will advertise as follows:

P(A)=\frac{n(A)}{N}=\frac{64}{120}=0.5333

Thus, the probability that next week the competitor will advertise is 0.5333.

(b)

Compute the probability that next week sales will not be high as follows:

P(H^{c})=1-P(H)=1-\frac{n(H)}{N}=1-\frac{41}{120}=\frac{120-41}{120}=0.6583

Thus, the probability that next week sales will not be high is 0.6583.

(c)

The events of achieving a medium or high sales are mutually exclusive.

Since the sales achieved will either be medium or high. They cannot be both.

So, P (M ∩ H) = 0.

Compute the probability that next week there will be medium or high sales will be achieved as follows:

P(M\cup H)=P(M)+P(H)=\frac{26}{120}+\frac{41}{120}=\frac{26+41}{120}=0.5583

Thus, the probability that next week there will be medium or high sales will be achieved is 0.5583.

(d)

Compute the probability that next week either the competitor will advertise, or only low sales will be achieved as follows:

P(A\cup L)=P(A)+P(L)-P(A\cap L)=\frac{64}{120}+\frac{53}{120}-\frac{32}{120}=\frac{85}{120}=0.7083

Thus, the  the probability that next week either the competitor will advertise, or only low sales will be achieved is 0.7083.

(e)

Compute the probability that next week either the competitor will not advertise, or high sales will be achieved as follows:

P(NA\cup H)=P(NA)+P(H)-P(NA\cap H)=\frac{56}{120}+\frac{41}{120}-\frac{23}{120}=\frac{74}{120}=0.6167

Thus, the  the probability that next week either the competitor will not advertise, or high sales will be achieved is 0.6167.

8 0
3 years ago
What will happenif you: multiply one charge by 2.0, multiply the other charge by 4.8, and multiply the distance between them by
noname [10]
Yes it is. 
F1 = k q1 * q2 / r^2
F2 = k *(2*q1) * (4.8 q2) / (7.2 r^2)

Work on F2 for a moment.
F2 = k * 9.6 (q1*q2) / 51.84
F2 = (9.6/ 51.84) *  k * q1 * q2 / r^2
Since k * q1 * q2/r^2 is the same in both questions let kq1*q2/r^2 = m

F1 = m
F2 = 0.18 m

So to make it easier F2/F1 = 0.18m / m
F2/ F1 = 0.18 the m's cancel.

And that should be how you do the question.
4 0
3 years ago
For the triangles described, which of the following statements must be true? In triangle DEF, DE=8 in., DF=23 in., and ∡D=16°. I
Temka [501]

Answer:

∠Q≅∠F

Step-by-step explanation:

Two triangles are said to be congruent if all the three sides of the triangles are equal and all the three angles are equal.

Given that: In triangle DEF, DE=8 in., DF=23 in., and ∡D=16°. In triangle PQR, PQ=23 in., PR=8 in., and ∡P=16°.

Hence we can say that ΔDEF is congruent to ΔPQR. According to the side-angle-side (SAS) triangle congruence theorem: If two sides and the included angle of one triangle are equal to two sides and the included angle of another triangle, the triangles are congruent.

Therefore:

DF = PQ, DE = PR, EF = RQ, ∠D = ∠P, ∠E = ∠R and ∠F = ∠Q

7 0
3 years ago
Factor by grouping <br><br> can someone please help me
Novosadov [1.4K]

Answer:

Re group then do your fraction

Step-by-step explanation:

7 0
3 years ago
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