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3 years ago

5 and 11Please Help!!!Help Is Appreciated!!!

Mathematics

2 answers:

posledela3 years ago

7 0

Do you have pictures or can your type in the problem??

disa [49]3 years ago

6 0

5+11? Or... 5*11? Or... 5/11? Or... 5-11? What do you need help with?

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Find the distance between the points (-3, 11) and (5, 5). 10 2 2

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Use the distance formula to find the distance between two points.

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

$d= \sqrt{(-3-5)^2+(11-5)^2} \\ d - \sqrt{-8^2+6^2} \\ d=\sqrt{64+36} \\ d = \sqrt{100} \\\\ \boxed{d=10}$

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PilotLPTM [1.2K]

The greatest common factor of 8 and 52 is 2.

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????? Anyone that can help giving 10 points

katen-ka-za [31]

Step-by-step explanation:

we know that perimeter of triangle is equals to side + side + side

but we can see that the particular triangle is having all the sides equal so it is seen to be equal lateral triangle and formula of equilateral triangle is 3 × sides.....

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3 years ago

The perimeter of equilateral triangle ABC is 81/3 centimeters, find the length of the radius and apothem.

MAXImum [283]

There is a typo error, the perimeter of equilateral triangle ABC is 81/√3 centimeters.

Answer:

Radius = OB= 27 cm

Apothem = 13.5 cm

A diagram is attached for reference.

Step-by-step explanation:

Given,

The perimeter of equilateral triangle ABC is 81/√3 centimeters.

Substituting this in the formula of perimeter of equilateral triangle = $3\times\ side$

$3\times\ side$ $=[tex]81\sqrt{3}$

$Side = \frac{81\sqrt{3} }{3} =27\sqrt{3} \ cm$

Thus from the diagram , Side $AB=BC=AC= 27\sqrt{3} \ cm$

We know each angle of an equilateral triangle is 60°.

From the diagram, OB is an angle bisector.

Thus $\angle OBC = 30$ °

Apothem is the line segment from the mid point of any side to the center the equilateral triangle.

Therefore considering ΔOBE, and applying tan function.

$tan\theta =\frac{perpendicular}{base} \\tan\theta=\frac{OE}{BE} \\tan\theta=\frac{OE}{\frac{27\sqrt{3}}{2} } \\tan30\times {\frac{27\sqrt{3} }{2} }= OE\\\frac{1}{\sqrt{3} } \times\frac{27\sqrt{3} }{2} =OE\\$

Thus ,apothem OE= 13.5 cm

Now for radius,

We consider ΔOBE

$cos\theta=\frac{base}{hypotenuse} \\cos30= \frac{BE}{OB} \\Cos30 = \frac{\frac{27\sqrt{3} }{2}}{OB} \\OB= \frac{\frac{27\sqrt{3} }{2}}{cos30} \\OB= \frac{\frac{27\sqrt{3} }{2}}{\frac{\sqrt{3} }{2} } \\OB =27 \ cm$