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2 years ago

A scientist cools some water at a constant rate. The graph and table show

Mathematics

1 answer:

Afina-wow [57]2 years ago

7 0

A function assigns the values. The function for this model can be written as y=-4x+50.

<h3>What is a Function?</h3>

A function assigns the value of each element of one set to the other specific element of another set.

As it can be seen from the graph that the function is linear, therefore, we can write,

y = mx + c

Given two points of the graph, therefore, we can write for the first point,

30 = m(5) + c

30 = 5m + c

c = 30 - 5m

Also, for the second point,

y = mx + c

26 = 6m + c

Substitute the value of c,

26 - 6m = 30 - 5m

-6m + 5m = 30 - 26

-m = 4

m = -4

Now, substitute the value of m in any one of the equation formed above,

c = 30 - 5m

c = 30 - 5(-4)

c = 50

Hence, the function for this model can be written as y=-4x+50.

Learn more about Function:

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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

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