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Ganezh [65]
3 years ago
5

Solve for X. Geometry

Mathematics
1 answer:
Dennis_Churaev [7]3 years ago
5 0

Answer:

x=12

Step-by-step explanation:

LM + MN = LN

2x-16 + x-9 = 11

Combine like terms

3x-25=11

Add 25 to each side

3x-25+25 = 11+25

3x = 36

Divide by 3

3x/3=36/3

x = 12

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Step-by-step explanation:

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8x-5y=-118x−5y=−118, x, minus, 5, y, equals, minus, 11 xxx-intercept: \Big((left parenthesis ,,comma \Big))right parenthesis yyy
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Read 2 more answers
Line L1 passes through points P(n,4) and Q(1,-2). Line L2 passes through R(4,3) and S(1,5). What is the value of n if the lines
Elena L [17]

Answer:

<h3>n = -8</h3>

Step-by-step explanation:

For two lines to be parallel, they must have the same slope

Get the slope of each line

For L1;

m1 = -2-4/1-n

m1 = -6/1-n

For L2;

m2 = 5-3/1-4

m2 = 2/-3

m2 = -2/3

Since they have the same slope, m1 = m2

-6/1-n = -2/3

6/1-n = 2/3

Cross multiply

2(1-n) = 6(3)

2-2n = 18

-2n = 18 - 2

-2n = 16

n = -16/2

n = -8

Hence the value of n is -8

4 0
2 years ago
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
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