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svlad2 [7]
3 years ago
7

Which of the following may fall outside a triangle? Check all that apply.?

Mathematics
1 answer:
Luda [366]3 years ago
5 0
Circumcenter is the one among the following choices given in the question that may fall outside a triangle. The correct option among all the options that are given in the question is the first option or option "A". I hope that this is the answer you were looking for and it has come to your help.
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I need your guys help please as soon as possible!
lesya [120]

Answer:

4(6)/2=12 + 4(4)/2= 8

12+8= 20 sq units

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3 years ago
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3.09 Geometry quuizzzzz
V125BC [204]
1. 90
2. 40
3. 14
4. 1
5. d
5 0
3 years ago
Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10
zvonat [6]

The approximate difference in the ages of the two cars, which  depreciate to 60% of their respective original values, is 1.7 years.

<h3>What is depreciation?</h3>

Depreciation is to decrease in the value of a product in a period of time. This can be given as,

FV=P\left(1-\dfrac{r}{100}\right)^n

Here, (<em>P</em>) is the price of the product, (<em>r</em>) is the rate of annual depreciation and (<em>n</em>) is the number of years.

Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10%.

Suppose the original price of the first car is x dollars. Thus, the depreciation price of the car is 0.6x. Let the number of year is n_1. Thus, by the above formula for the first car,

0.6x=x\left(1-\dfrac{10}{100}\right)^{n_1}\\0.6=(1-0.1)^{n_1}\\0.6=(0.9)^{n_1}

Take log both the sides as,

\log 0.6=\log (0.9)^{n_1}\\\log 0.6={n_1}\log (0.9)\\n_1=\dfrac{\log 0.6}{\log 0.9}\\n_1\approx4.85

Now, the second car depreciates at an annual rate of 15%. Suppose the original price of the second car is y dollars.

Thus, the depreciation price of the car is 0.6y. Let the number of year is n_2. Thus, by the above formula for the second car,

0.6y=y\left(1-\dfrac{15}{100}\right)^{n_2}\\0.6=(1-0.15)^{n_2}\\0.6=(0.85)^{n_2}

Take log both the sides as,

\log 0.6=\log (0.85)^{n_2}\\\log 0.6={n_2}\log (0.85)\\n_2=\dfrac{\log 0.6}{\log 0.85}\\n_2\approx3.14

The difference in the ages of the two cars is,

d=4.85-3.14\\d=1.71\rm years

Thus, the approximate difference in the ages of the two cars, which  depreciate to 60% of their respective original values, is 1.7 years.

Learn more about the depreciation here;

brainly.com/question/25297296

4 0
2 years ago
An equilateral is shown inside a square inside a regular pentagon inside a regular hexagon. The square and regular hexagon are s
Kay [80]

Shaded area = area of the hexagon – area of the pentagon + area of the square – area of the equilateral triangle. This can be obtained by finding each shaded area and then adding them.

<h3>Find the expression for the area of the shaded regions:</h3>

From the question we can say that the Hexagon has three shapes inside it,

  • Pentagon
  • Square
  • Triangle

Also it is given that,

An equilateral triangle is shown inside a square inside a regular pentagon inside a regular hexagon.

From this we know that equilateral triangle is the smallest, then square, then regular pentagon and then a regular hexagon.

A pentagon is shown inside a regular hexagon.

  • Area of first shaded region = Area of the hexagon - Area of pentagon

An equilateral triangle is shown inside a square.

  • Area of second shaded region = Area of the square - Area of equilateral triangle  

The expression for total shaded region would be written as,

Shaded area = Area of first shaded region + Area of second shaded region

Hence,        

⇒ Shaded area  = area of the hexagon – area of the pentagon + area of the  square – area of the equilateral triangle.

 

Learn more about area of a shape here:

brainly.com/question/16501078

#SPJ1

8 0
1 year ago
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Someone help me with this proof please ​
Anna71 [15]

Answer:

the first statement is: AB is congruent to DC.

reason: given

Step-by-step explanation:

im not sure about the others

6 0
3 years ago
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