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2 years ago

Which dimensions can create only one unique triangle?

1 answer:

4 0

The answer to this question has been answered already, just type the question into the search to see if someone ask the same question already.

case A. Three angles measuring 25,65,and 90

sum of a internal angles=25+65+90-----> 180° is ok

That dimensions can create infinity similar triangles

case B. Three angles measuring 50,50,and 50

sum of a internal angles=50+50+50----> 150° is not a triangle

therefore

That dimensions cannot create a triangle

case C. Three sides measuring 5 in,12 in,and 14 in

Applying the Triangle Inequality Theorem

5+12>14 -- > is ok

14+5>12 --- > is ok

12+14>5 --- > is ok

That dimensions can create only one unique triangle

case D. Three sides measuring 4 ft,8 ft, and 14 ft

Applying the Triangle Inequality Theorem

4+8>14 -- > is not true

therefore

That dimensions cannot create a triangle

the answer is

Option C. Three sides measuring 5 in,12 in,and 14 in

Read more on Brainly.com - brainly.com/question/9930431#readmore

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2 years ago

What number is 2 * 10 ^100

Studentka2010 [4]

Answer:

2.00000000e+100

Step-by-step explanation:

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2 years ago

Prove that the roots of x2+(1-k)x+k-3=0 are real for all real values of k

masha68 [24]

Answer:

Roots are not real

Step-by-step explanation:

To prove : The roots of x^2 +(1-k)x+k-3=0x

+(1−k)x+k−3=0 are real for all real values of k ?

Solution :

The roots are real when discriminant is greater than equal to zero.

i.e. b^2-4ac\geq 0b

−4ac≥0

The quadratic equation x^2 +(1-k)x+k-3=0x

+(1−k)x+k−3=0

Here, a=1, b=1-k and c=k-3

Substitute the values,

We find the discriminant,

D=(1-k)^2-4(1)(k-3)D=(1−k)

−4(1)(k−3)

D=1+k^2-2k-4k+12D=1+k

−2k−4k+12

D=k^2-6k+13D=k

−6k+13

D=(k-(3+2i))(k+(3+2i))D=(k−(3+2i))(k+(3+2i))

For roots to be real, D ≥ 0

But the roots are imaginary therefore the roots of the given equation are not real for any value of k.

6 0

2 years ago

Thank you very much anh you

Neko [114]

Hello from MrBillDoesMath!

Answer:

See discussion below

Discussion:

The sum of the interior angles of a polygon of n sides is (n-2)* 180. As the polygon in #4 has 7 sides , the sum of the interior angles is (7-2)*180 = 5 * 180 = 900.

We can also compute the sum of the interior angles by adding up each of the angles (expressed in terms of x) shown in #4. Starting at the top of the polygon and proceeding counterclockwise gives:

900 = (8x +34) + (10x+21)+(9x+30)+(7x+45)+(5x+44)+(12x+13)+(6x+29) =>

(8x + 10x+9x+7x+5x+12x+6x) + (34+21+30+45+44+13+29) =

57x + 216

This simplifies to

900 = 57x + 216 => subtract 216 from both sides

900-216 = 57x => 900-216= 684

684= 57x => divide both sides by 57

684/57 = x => as 684/57 = 12

x = 12

The diagram has makes repeated use of the same variable for vertices. That is, the diagram shows 4 P's, 1 Q, 2 R's, 0 S's, and 1 T making it impossible to determine the values of some angles. For example, consider m RST -- whatever that means.

#5 uses the same ideas but the vertices are properly labeled (I'll leave the grunt of determining the individual angles to you but here are the main points).

Sum of interior angles = (7 -2) * 180 = 5 * 180 = 900.

900 = (8x+62) + (7x+66)+(5x+83)+(6x+55)+(10x+42)+(8x+45)+(4x+67) =>

900 = (8x + 7x +5x+6x+10x+8x+4x) + (62+66+83+55+42+45+67)

900= 48x + 420 =>

900 -420 = 48x => subtract 420

480 = 48x => divide by 48

x = 480/48 = 10

Thank you,

MrB

6 0

3 years ago