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abruzzese [7]
3 years ago
13

WILL GIVE A BRAINLEST

Mathematics
2 answers:
Gnoma [55]3 years ago
8 0

Answer: "No, the triangles are not necessarily congruent." is the correct statement .


Step-by-step explanation:

In ΔCDE, m∠C = 30° and m∠E = 50°

Therefore by angle sum property of triangles

m∠C+m∠D+m∠E=180°

⇒m∠D=180°-m∠E-m∠C=180°-30°-50°=100°

⇒m∠D=100°

In ΔFGH, m∠G = 100° and m∠H = 50°

Similarly m∠F +∠G+m∠H=180°

⇒m∠F=180°-∠G-m∠H=180°-100°-50=30°

⇒m∠F=30°

Now ΔCDE and ΔFGH

m∠C=m∠F=30°,m∠D=m∠G=100°,m∠E=m∠H=50°

by AAA similarity criteria  ΔCDE ≈ ΔFGH but can't say congruent.

Congruent triangles are the pair of triangles in which corresponding sides and angles are equal . A congruent triangle is a similar triangle but a similar triangle may not be a congruent triangle.


rusak2 [61]3 years ago
8 0

Answer:

No, the triangles are not necessarily congruent.

Its D on EDGE

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3 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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