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3 years ago

Suppose f and g are continuous functions such that g(2) = 6 and lim x → 2 [3f(x) + f(x)g(x)] = 36. find f(2).

Mathematics

1 answer:

True [87]3 years ago

7 0

Answer: f(2) = 4

Step-by-step explanation:

F(x) and g(x) are said to be continuous functions

Lim x=2 [3f(x) + f(x)g(x)] = 36

g(x) = 2

Limit x=2

[3f(2) + f(2)g(2)] = 36

[3f(2) + f(2) . 6] = 36

[3f(2) + 6f(2)] = 36

9f(2) = 36

Divide both sides by 9

f(2) = 36/9

f(2) = 4

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Please help?

Mazyrski [523]

Answer:

$23\sqrt{3}\ un^2$

Step-by-step explanation:

Connect points I and K, K and M, M and I.

1. Find the area of triangles IJK, KLM and MNI:

$A_{\triangle IJK}=\dfrac{1}{2}\cdot IJ\cdot JK\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 2\cdot 3\cdot \dfrac{\sqrt{3}}{2}=\dfrac{3\sqrt{3}}{2}\ un^2\\ \\ \\A_{\triangle KLM}=\dfrac{1}{2}\cdot KL\cdot LM\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 8\cdot 2\cdot \dfrac{\sqrt{3}}{2}=4\sqrt{3}\ un^2\\ \\ \\A_{\triangle MNI}=\dfrac{1}{2}\cdot MN\cdot NI\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 3\cdot 8\cdot \dfrac{\sqrt{3}}{2}=6\sqrt{3}\ un^2\\ \\ \\$

2. Note that

$A_{\triangle IJK}=A_{\triangle IAK}=\dfrac{3\sqrt{3}}{2}\ un^2 \\ \\ \\A_{\triangle KLM}=A_{\triangle KAM}=4\sqrt{3}\ un^2 \\ \\ \\A_{\triangle MNI}=A_{\triangle MAI}=6\sqrt{3}\ un^2$

3. The area of hexagon IJKLMN is the sum of the area of all triangles:

$A_{IJKLMN}=2\cdot \left(\dfrac{3\sqrt{3}}{2}+4\sqrt{3}+6\sqrt{3}\right)=23\sqrt{3}\ un^2$

Another way to solve is to find the area of triangle KIM be Heorn's fomula, where all sides KI, KM and IM can be calculated using cosine theorem.

7 0

3 years ago

What is the classification of the polynomial by the number of terms?

hichkok12 [17]

Answer: Trinomial

Step-by-step explanation:

8 0

3 years ago

What is the value of this expression? 100 - [(5 x 3 + 4)]

shepuryov [24]

Answer:

Step-by-step explanation:

Order of operations rules require that we do the work inside parentheses first. Multiplication must be done before addition. Thus, the quantity inside parentheses becomes 15 + 4 = 19.

Next, we evaluate 100 - [19], obtaining 81 (answer)

7 0

3 years ago

Read 2 more answers

The lengths, in minutes, of the movies currently showing at a movie theater are shown in the data set. Create a histogram

alexdok [17]

Answer:

See Explanation

Step-by-step explanation:

The question is incomplete, as the required lengths are not given.

I will use the following data set to answer the question.

$Dataset: 35, 51, 39, 42, 68, 62, 32, 59$

First, is to determine the range of the dataset

$Range = Highest - Least$

$Range = 68 - 32$

$Range = 36$

Next, we will make use of 4 classes. So, we divide range by 10 to get the number of class. 10 represents the interval

$Class= Range/10$

$Class = 36/10$

$Class = 3.6$

$Class \approx 4$

<em>So, we use 4 classes</em>

Plot the frequency distribution table as follows:

$\begin{array}{ccc}{Intervals} & {Observations} & {Frequency} & {31-40} & {32,35,39} & {3} & {41-50} & {42} & {1} & {51-60} & {51,59} & {2} & {61-70} & {62,68} & {2}\ \end{array}$

<em>See attachment for histogram</em>

5 0

3 years ago

The new Fore and Aft Marina is to be located on the Ohio River near Madison, Indiana. Assume that Fore and Aft decides to build

RUDIKE [14]

Answer:

A.) 0.2

B.) 3.2

C.) 0.4

D.) 0.5

Step-by-step explanation:

Arrival rate, λ = 8 per hour

Service rate, μ = 10 boats per hour

1.)

Probability that boat dock will be idle, P0:

P0 = 1 - λ/μ

P0 = 1 - 8/10

P0 = 1 - 0.8

P0 = 0.2

Average number of boats waiting for service :

Lq = λ² ÷ μ(μ - λ)

Lq = 8² ÷ 10(10- 8)

Lq = 64 ÷ 10(2)

Lq = 64 ÷ 20

Lq = 3.2

Average time spent waiting for service, Wq :

Wq = λ ÷ μ(μ - λ)

Wq = 8 ÷ 10(10- 8)

Wq = 8 ÷ 10(2)

Wq = 8 / 20

Wq = 0.4 hour

Wq = 0.4 * 60 = 24 minutes

Average time spent at the dock :

1 ÷ (μ - λ)

1 ÷ (10 - 8)

1 ÷ 2

= 0.5 hour

= 0.5 * 60 = 30 minutes

4 0

3 years ago