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4 years ago

Examine the function f(x)=x+(3/x). Find the point on the curve at which the tangent lines pass through the point (1, 1).

1 answer:

6 0

Base on the function that you give and the data that are given. The point on the curve at which the tangent lines pass through the point (1,1). Base on my calculation and through my analyzations i came up with an answer of -2x+3 = x+3/x

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What is the measure of angle c to the nearest whole degree?

iris [78.8K]

We can use a trigonometric function to solve for angle C. We can use any of the trigonometric functions to solve, but today let's use tangent.

Since we know the dimensions of the opposite and adjacent legs, we can solve for theta directly:

$tan(theta)=\frac{opp}{adj}$

$tan(theta)=\frac{24}{45}$

$theta=tan^{-1}(0.5333)$

$theta=0.4899573$ - but this is in radians. So we must convert to degrees.

$Degrees=\frac{180}{\pi}*radians$

$Degrees=\frac{180}{\pi}*0.4899573$

$Degrees=28.07$

Therefore, the measure of angle C is 28°.

6 0

4 years ago

A wire of length L is cut at the red point. The segment to the left of the cut is formed into an equilateral triangle, and the s

Elden [556K]

Answer:

$(a) x=0\\(b)\ x=\dfrac{4L}{4+\sqrt{3}}$

Step-by-step explanation:

Imagine that the wire is cut as shown in the picture. We know that the lenght of the segment to the left is $x$ and therefore the lenght of the segment to the right must be $L-x.$

Using the formula for the area of an equilateral triangle of sides of length x, we get that $A_{\bigtriangleup}(x)=\dfrac{\sqrt{3}}{4}\, x^2.$

And using the formula for the area of a square of length L-x, we obtain that $A_{\square}(x)=(L-x)^2.$

Then, the total area of the two shapes is giving by the sum of both areas: $A_T(x)=A_{\bigtriangleup}(x)+A_{\square}(x)=\dfrac{\sqrt{3}}{4}\, x^2 + (L-x)^2.$

Now we have to find the values x where the function $A_T(x)$ attains its maximum and its minimum. For this purpose, we calculate its critical points, which occurs when the derivative vanishes:

$A_T'(x)=A_{\bigtriangleup}'(x)+A_{\square}'(x)=\dfrac{\sqrt{3}}{2}x-2(L-x)=0.$

Solving for x we get that:

$x=\dfrac{4L}{4+\sqrt{3}}.$

This is the only critical point. Using the second derivative test we found that, since $A_T''(x)=\dfrac{1}{2} \left(4 + \sqrt{3}\right) >0,$ then at $x=\dfrac{4L}{4+\sqrt{3}}$ the area of the two shapes is minimized.

Now, the only way we have maximum area is when the red point is on the extreme of the wire, at x = 0. This because in that situation, there is no triangle that can be formed and therefore the area is equals $L^2.$

4 0

4 years ago

1) Given the area of a Football field (57,600 sqft), and the dimensions of a

Neporo4naja [7]

Answer:

Option (d).

Step-by-step explanation:

Given that,

The area of a Football field is 57,600 sqft

The dimensions of Basketball court is 84ft x 50ft

We need to find what fraction of the football crowd would fit on the court. Taking ratio of the area of basket ball court to the football court as follows :

$R=\dfrac{84\times 50}{57,600}\\\\R=\dfrac{7}{96}\ =\ 0.0729$

It is less than a tenth. Hence, the correct option is (d).

3 0

3 years ago

Suppose it is known that the individual lost more than 12 pounds in a month. Find the probability that he lost less than 19 poun

telo118 [61]

The probability that he lost less than 19 pounds is 7 / 9.

According to statement

The weight loss by individual is 12 Pounds

So, by uniform distribution probability

P(c ≤ x ≤ d) = (d - c) / (b - a)

Substitute the values in it then

P(12 ≤ x ≤ 19) = (19 - 12) / (20 - 11)

P(12 ≤ x ≤ 19) = 7 / 9

So, the probability that he lost less than 19 pounds is 7 / 9.

Learn more about UNIFORM DISTRIBUTION PROBABILITY here brainly.com/question/14114556

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5 0

2 years ago

(7,3) a solution of y

nasty-shy [4]

Answer:

y=3

Step-by-step explanation:

3 0

3 years ago