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Crank
3 years ago
13

How many possible outcomes are there when flipping a coin nine times

Mathematics
1 answer:
Vedmedyk [2.9K]3 years ago
6 0
18 is my best answer

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Intersection point of Y=logx and y=1/2log(x+1)
GalinKa [24]

Answer:

The intersection is (\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2}).

The Problem:

What is the intersection point of y=\log(x) and y=\frac{1}{2}\log(x+1)?

Step-by-step explanation:

To find the intersection of y=\log(x) and y=\frac{1}{2}\log(x+1), we will need to find when they have a common point; when their x and y are the same.

Let's start with setting the y's equal to find those x's for which the y's are the same.

\log(x)=\frac{1}{2}\log(x+1)

By power rule:

\log(x)=\log((x+1)^\frac{1}{2})

Since \log(u)=\log(v) implies u=v:

x=(x+1)^\frac{1}{2}

Squaring both sides to get rid of the fraction exponent:

x^2=x+1

This is a quadratic equation.

Subtract (x+1) on both sides:

x^2-(x+1)=0

x^2-x-1=0

Comparing this to ax^2+bx+c=0 we see the following:

a=1

b=-1

c=-1

Let's plug them into the quadratic formula:

x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

x=\frac{1 \pm \sqrt{(-1)^2-4(1)(-1)}}{2(1)}

x=\frac{1 \pm \sqrt{1+4}}{2}

x=\frac{1 \pm \sqrt{5}}{2}

So we have the solutions to the quadratic equation are:

x=\frac{1+\sqrt{5}}{2} or x=\frac{1-\sqrt{5}}{2}.

The second solution definitely gives at least one of the logarithm equation problems.

Example: \log(x) has problems when x \le 0 and so the second solution is a problem.

So the x where the equations intersect is at x=\frac{1+\sqrt{5}}{2}.

Let's find the y-coordinate.

You may use either equation.

I choose y=\log(x).

y=\log(\frac{1+\sqrt{5}}{2})

The intersection is (\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2}).

6 0
3 years ago
A fishing lodge wants to build a new boat ramp. The slope of the ramp must be 8° to the horizontal. The total height of the bank
Likurg_2 [28]

Answer:

The length of the sloping section of the ramp is 20.12 m

Step-by-step explanation:

Given;

the total height of the bank, h = 2.8 m

The slope of the ramp must be 8° to the horizontal, i.e, θ = 8°

Let the length of the sloping section = L

let the horizontal distance between the height of the bank and sloping section = b

Thus, h, L and b forms three sides of a right angled-triangle, with L as the hypotenuse side, h  (height of the triangle) as the opposite side and b  (base of the triangle) as the adjacent side.

We determine L by applying the following formula;

Sinθ = opposite / hypotenuse

Sin θ = h / L

L = h / Sin θ

L = 2.8 / Sin 8

L = 2.8 / 0.13917

L = 20.12 m

Therefore, the length of the sloping section of the ramp is 20.12 m

6 0
3 years ago
Can you solve this? :)
Nina [5.8K]
The answer is B.
X = -11 :-)
8 0
3 years ago
Read 2 more answers
Need help asap for geometry​
kupik [55]

Answer:

35

Step-by-step explanation:

40=1/2(115-x)

80=115-x

-35=-x

x=35

4 0
3 years ago
Read 2 more answers
Directions in picture.
densk [106]
Use PEMDAS, the answer that I got is 19.
8 0
3 years ago
Read 2 more answers
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