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lutik1710 [3]
3 years ago
14

1. (x + 2) (x +5)

Mathematics
1 answer:
madam [21]3 years ago
3 0

Step-by-step explanation:

el trabajo que pusiste es muy falso

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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
Solve solve the following equations 2x (x minus 5) close + 3(x-2)=8+7(x-4)​
stiv31 [10]

Step-by-step explanation:

<em>2</em><em>x</em><em>(</em><em>x-5</em><em>)</em><em>+</em><em>3</em><em>(</em><em>x-2</em><em>)</em><em>=</em><em>8</em><em>+</em><em>7</em><em>(</em><em>x-4</em><em>)</em>

<em>2</em><em>x</em><em>²</em><em>-10x</em><em>+</em><em>3</em><em>x</em><em>-</em><em>6</em><em>=</em><em>8</em><em>+</em><em>7</em><em>x</em><em>-</em><em>2</em><em>8</em><em>(</em><em>Group</em><em> </em><em>like</em><em> </em><em>terms</em><em>)</em>

<em>2</em><em>x</em><em>²</em><em>-</em><em>1</em><em>0</em><em>x</em><em>+</em><em>3</em><em>x</em><em>-</em><em>7</em><em>x</em><em>=</em><em>8</em><em>-</em><em>2</em><em>8</em><em>+</em><em>6</em>

<em>2</em><em>x</em><em>²</em><em>-</em><em>7</em><em>x</em><em>-</em><em>7</em><em>x</em><em>=</em><em>-</em><em>2</em><em>0</em><em>+</em><em>6</em>

<em>2</em><em>x</em><em>²</em><em>-</em><em>1</em><em>4</em><em>x</em><em>=</em><em>-</em><em>1</em><em>4</em><em>(</em><em>Divi</em><em>de</em><em> </em><em>both</em><em> </em><em>sides</em><em> </em><em>by</em><em> </em><em>-</em><em>1</em><em>4</em><em>)</em>

2x²<em>-</em><em>x</em><em>=</em><em>1</em>

<em>2</em><em>x</em><em>=</em><em>1</em>

<em>x</em><em>=</em><em>½</em>

<em>Hope</em><em> </em><em>you </em><em>get</em><em> </em><em>it</em><em> </em><em>please</em><em> </em><em>someone</em><em> </em><em>shou</em><em>ld</em><em> </em><em>verify</em><em>.</em><em>.</em><em>.</em><em>.</em>

7 0
3 years ago
I NEED HELP PLEASE, THANKS! :)
KIM [24]

Answer:  tan(14x)

<u>Step-by-step explanation:</u>

Consider the Sum Formula for tan:

tan(A + B)=\dfrac{tan(A)+tan(B)}{1-tan(A)(tan(B)}\\\\\\tan(9x+5x)=\dfrac{tan(9x)+tan(5x)}{1-tan(9x)(tan(5x)}\\\\\\\large\boxed{tan(14x)}=\dfrac{tan(9x)+tan(5x)}{1-tan(9x)(tan(5x)}

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