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2 years ago

What is HL??? HL=__ Image Below

Mathematics

1 answer:

Advocard [28]2 years ago

4 0

Consider a trapezoid GJKF with:

GH=JH (hypothesis)

FL=KL (hypothesis)

so HL is the median of trapezoid GJKF

so: HL=1/2(2.5+1.2)=1.85

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Each earbud is 4/7 of an inch long. If you put 8 earbuds end to end, how long would they be from beginning to end?

Lisa [10]

Answer:

4 4/7

Step-by-step explanation:

4/7x8/1

4x8/7x1

32/7

4 4/7

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2 years ago

Read 2 more answers

57:22 5 What is the equation, in point-slope form, of the line that is perpendicular to the given line and passes through the po

steposvetlana [31]

Answer:

C. y + 3 = ¼(x + 4)

Step-by-step explanation:

✔️Find the slope of the given line:

Slope = ∆y/∆x = -(4/1) = -4

The line that is perpendicular to the given line on the graph would have a slope that is the negative reciprocal of -4.

Thus, the slope of the line that is perpendicular to the line on the graph would be ¼.

m = ¼.

Since the line passes through (-4, -3), to write the equation in point-slope form, substitute a = -4, b = -3, and m = ¼ into y - b = m(x - a)

Thus:

y - (-3) = ¼(x - (-4))

y + 3 = ¼(x + 4)

7 0

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

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