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2 years ago

Which of the following sets of numbers could not represent the three sides of a triangle?

Mathematics

1 answer:

timama [110]2 years ago

8 0

Answer:

Step-by-step explanation:

In order for the sides lengths to form a triangle, the sum of the shortest two must exceed the longest.

That will not be the case for ...

{13, 16, 30}

because 13 +16 = 29 < 30.

(13, 16, 30} could not form a triangle

For more on the subject, see ...

brainly.com/question/26258903

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What is 0.77 estimated

galina1969 [7]

The answer would be .80 or it may be 1

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2 years ago

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Refer to the figure below. If m<60 degrees l, find the measure of the other angles. Show your work(or explain) to receive ful

gladu [14]

All of the angles combined =360 degrees

Vertical angles are congruent, so if angle 1 = 60, so does angle 3

Logically we have to find the measure of angle 4 and angle 2 combined and divide by 2

60 + 60 = 120

360 - 120 = 240

240 / 2 = 120

angle 4 = 120
angle 2= 120

angle 3 = 60

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3 years ago

Based on the results shown in the table, how many white marbles would one expect to draw in a total of 50 draws? (hint:set up a

Dovator [93]

The answer to your question is 10

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3 years ago

If D is midpoint of side BC of triangle ABC. P and Q are points lying respectively on side AB and AC such that DP is parallel to

LUCKY_DIMON [66]

See proof below

Step-by-step explanation:

Assume triangle ABC to have vertices at;

A(2,-1), B(2,-7) and C(6,-7)

D is midpoint of BC, thus D is at (4,-7)

The P and Q, are lying on side AB and AC, hence assume P is at (2,-4) and Q is at (4,-4) such at DP is parallel to QA

Plot the points on a graph tool and join the points to view the sketch.

To prove area of triangle CPQ is 1/4 area of ABC will be;

Find area ABC and CPQ then compare the areas.

Apply the distance formula to find the length of sides of the triangles then find the areas.

The distance formula is;

$d=\sqrt{x_2-x_1)^2+(y_2-y_1)^2}$

Length of side AB from the sketch is;

$AB=\sqrt{(2-2)^2+(-7--1)^2} =\sqrt{-6^2} =\sqrt{36} =6units$

Length of side BC will be;

$BC=\sqrt{(-7--7)^2+(6-2)^2} =\sqrt{4^2} =\sqrt{16} =4units$

Thus area of triangle ABC will be;

1/2 *base length*height ------because it is a right-triangle

1/2*4*6=12 square units

Find the lengths of all sides of triangle CPQ

Length of side PQ is half that of side BC thus PQ=2 units

Length of side PC is;

$PC=\sqrt{(6-2)^2+(-7--4)^2} =\sqrt{4^2+-3^2} =\sqrt{16+9} =\sqrt{25} =5units$

Length of side QC will be;

$QC=\sqrt{(6-4)^2+(-7--4)^2} =\sqrt{2^2+-3^3} =\sqrt{4+9} =\sqrt{13}$

QC= √13 = 3.6 units

Find area of triangle CPQ given all sides by applying the Heron's formula for area of triangle which is;

A=√s(s-a)(s-b)(s-c) where;

A=area of the triangle

s= half the perimeter of the triangle

a=side PQ = 2 units

b=side PC = 5 units

c= side QC = 3.6 units

Finding the perimeter of triangle CPQ will be;

P=sum of all sides

P=2+5+3.6 =10.6 units

s=10.6/2 = 5.3

Area of the triangle CPQ will be;

$A=\sqrt{5.3(5.3-2)(5.3-5)(5.3-3.6)} \\A=\sqrt{5.3(3.3)(0.3)(1.7)} \\A=\sqrt{8.9} =2.98$

A=3.0 (1 decimal place)

Compare the areas;

Area of triangle ABC=12 square units

Area of triangle CPQ = 3 square units

Area of triangle CPQ / Area of triangle ABC = 3/12 =1/4

Thus you have proved that area of triangle CPQ is 1/4 th area of triangle ABC because 1/4 *12 =3

Learn More

Area of a triangle ;brainly.com/question/14869984

The Heron's formula : brainly.com/question/10713495

Keywords: midpoint, triangle, sides, parallel, prove , area, equal

#LearnwithBrainly

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2 years ago

Please help me with questions 13-22!! I know I already answered 14 but if it is wrong please tell me!!! I don't have much time!!

neonofarm [45]

Answer:

13. no solution

15. 3

16. 2

17. 5

18. -8

19. -7

20. 6

21. -5

22. -7

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3 years ago