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allochka39001 [22]
3 years ago
13

When measuring the sides, triangles can be classified as acute, scalene and isoceles

Mathematics
1 answer:
ser-zykov [4K]3 years ago
3 0

Answer:

Yes, When measuring the sides, triangles can be classified as acute, scalene, and isosceles

Step-by-step explanation:

In any triangle, there are at least two acute angles

The types of triangles according to their angles are:

  • Acute-angled triangle ⇒ 3 angles are acute
  • Right-angled triangle ⇒ 2 acute angles and 1 right angle
  • Obtuse-angled triangle ⇒ 2 acute angles and 1 obtuse angle

The types of triangles according to their sides are:

  • Equilateral triangle ⇒ 3 equal sides
  • Isosceles triangle ⇒ 2 equal sides
  • Scalene triangle ⇒ 3 different side

<em>You can classify the type of the triangle according to its angle using the measuring of its sides.</em>

If a triangle has sides length a, b, c where c is the longest side

∵ a² + b² > c²

∴ The triangle is an acute-angled triangle

∵ a² + b² = c²

∴ The triangle is a right-angled triangle

∵ a² + b² < c²

∴ The triangle is an obtuse-angled triangle

<em>You can classify the type of the triangle according to its sides using the measuring of its sides.</em>

If a triangle has sides lengths a, b, c

∵ a, b, c have unequal lengths

∴ The triangle is scalene

∵ a and b or a and c or b and c have equal lengths

∴ The triangle is Isosceles

∵ a, b, c have equal lengths

∴ The triangle is equilateral

Yes, When measuring the sides, triangles can be classified as acute, scalene, and isosceles

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Answer:

Part a) About 48.6 feet

Part b) About 8.3 feet

Part c) The domain is 0 \leq x \leq 48.6\ ft and the range is 0 \leq y \leq 8.3\ ft

Step-by-step explanation:

we have

y=-0.014x^{2} +0.68x

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

where

x is the ball's  distance from the catapult in feet

y is the flight of the balls in feet

Part a)  How far did the ball fly?

Find the x-intercepts or the roots of the quadratic equation

Remember that

The x-intercept is the value of x when the value of y is equal to zero

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

in this problem we have

-0.014x^{2} +0.68x=0

so

a=-0.014\\b=0.68\\c=0

substitute in the formula

x=\frac{-0.68(+/-)\sqrt{0.68^{2}-4(-0.014)(0)}} {2(-0.014)}

x=\frac{-0.68(+/-)0.68} {(-0.028)}

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Part b) How high above the ground did the ball fly?

Find the maximum (vertex)

y=-0.014x^{2} +0.68x

Find out the derivative and equate to zero

0=-0.028x +0.68

Solve for x

0.028x=0.68

x=24.3

<em>Alternative method</em>

To determine the x-coordinate of the vertex, find out the midpoint  between the x-intercepts

x=(0+48.6)/2=24.3\ ft

To determine the y-coordinate of the vertex substitute the value of x in the quadratic equation and solve for y

y=-0.014(24.3)^{2} +0.68(24.3)

y=8.3\ ft

the vertex is the point (24.3,8.3)

therefore

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Part c) What is a reasonable domain and range for this function?

we know that

A  reasonable domain is the distance between the two x-intercepts

so

0 \leq x \leq 48.6\ ft

All real numbers greater than or equal to 0 feet and less than or equal to 48.6 feet

A  reasonable range is all real numbers greater than or equal to zero and less than or equal to the y-coordinate of the vertex

so

we have the interval -----> [0,8.3]

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