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Snezhnost [94]
2 years ago
5

Find the missing side of the triangle

Mathematics
2 answers:
Vilka [71]2 years ago
5 0

Answer:

the missing side of triangle is

\sqrt{y {}^{2} - {26}^{2}}

masya89 [10]2 years ago
3 0

Answer:

  • y = 66.66

Step-by-step explanation:

<u>Use trigonometry:</u>

  • cosine = adjacent / hypotenuse
  • cos 67° = 26/y
  • y = 26 / cos 67°
  • y = 26 / 0.39
  • y = 66.66
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Answer:

50.00 + 40.00

es 90.00

4 0
3 years ago
What is number 5???!!
vagabundo [1.1K]

The value of m∠M is 112°

Step-by-step explanation:

In the parallelogram NPMQ, opposite angles ∠P=∠M.

To get the value of the angles, equate the two expressions;

(6x-2)°=(4x+36)°

6x-2=4x+36 ------ collect like terms

6x-4x=36+2

2x=38

x=38/2 =19

So, ∠M= (6x-2)° = (6×19 - 2)° = 112°

Learn More

Angles in a parallelogram : brainly.com/question/11611093

Keyword : angles

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8 0
3 years ago
What is the slope ?simplify your answer and write it as a proper fraction ,improper fraction or interfere.
34kurt

Answer:

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4 0
3 years ago
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HELP PLEASE HELP!!!!!
babunello [35]

Answer:

(i): A triangle where all three sides are equal and the same.

(ii): A triangle where all three sides are unequal in length.

(iii): A triangle that has two sides of equal length.

(iv): A triangle where all three internal angles are acute(they measure less than 90 degrees.

(v): A triangle where all three internal angles are obtuse(measure above 90 degrees).

(vi): A triangle where one internal angle is a right angle(90 degrees).

Hope this helps

Step-by-step explanation:

6 0
2 years ago
He vertices of square pqrs are p -4,0 q 4,3 r 7,-5 and s -1,-18.Show that the diagonals of square pqrs are congruent perpendicul
Anit [1.1K]

Answer:

Step-by-step explanation:

The vertices of the square given are P(-4, 0), Q(4, 3), R(7, -5) and, S(-1, -18)

For this diagonal to be right angle the slope of the diagonal must be m1=-1/m2

So let find the slope of diagonal 1

The two points are P and R

P(-4, 0), R(7, -5)

Slope is given as

m1=∆y/∆x

m1=(y2-y1)/(x2-x1)

m1=-5-0/7--4

m1=-5/7+4

m1=-5/11

Slope of the second diagonal

Which is Q and S

Q(4, 3), S(-1, -18)

m2=∆y/∆x

m2=(y2-y1)/(x2-x1)

m2=(-18-3)/(-1-4)

m2=-21/-5

m2=21/5

So, slope of diagonal 1 is not equal to slope two

This shows that the diagonal of the square are not diagonal.

But the diagonal of a square should be perpendicular, this shows that this is not a square, let prove that with distance between two points

Given two points

(x1,y1) and (x2,y2)

Distance between the two points is

D=√(y2-y1)²+(x2-x1)²

For line PQ

P(-4, 0), Q(4, 3)

PQ=√(3-0)²+(4--4)²

PQ=√(3)²+(4+4)²

PQ=√9+8²

PQ=√9+64

PQ=√73

Also let fine RS

R(7, -5) and, S(-1, -18)

RS=√(-18--5)+(-1-7)

RS=√(-18+5)²+(-1-7)²

RS=√(-13)²+(-8)²

RS=√169+64

RS=√233

Since RS is not equal to PQ then this is not a square, a square is suppose to have equal sides

But I suspect one of the vertices is wrong, vertices S it should have been (-1,-8) and not (-1,-18)

So using S(-1,-8)

Let apply this to the slope

Q(4, 3), S(-1, -8)

m2=∆y/∆x

m2=(y2-y1)/(x2-x1)

m2=(-8-3)/(-1-4)

m2=-11/-5

m2=11/5

Now,

Let find the negative reciprocal of m2

Reciprocal of m2 is 5/11

Then negative of it is -5/11

Which is equal to m1

Then, the square diagonal is perpendicular

6 0
3 years ago
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