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2 years ago

How many unique triangles can be drawn with side lengths 8 in., 12 in., and 24 in.? Explain

Mathematics

1 answer:

Debora [2.8K]2 years ago

5 0

By using the triangular inequality, we will see that no triangles can be made with these side lengths.

<h3>How many triangles can be made with these side lengths?</h3>

Remember that for any triangle with side lengths A, B, and C, the triangular inequality must be true.

This means that the sum of any two sides must be larger than the other side.

A + B > C

A + C > B

B + C > A.

For the given side lengths, we will have:

8 in + 12 in > 24 in

8in + 24 in > 12 in

12 in + 24 in > 8 in.

Now, notice that the first inequality is false. So the triangular inequality is not meet. Then we can't make a triangle with these side lengths.

So we can make 0 unique triangles with these side lengths.

If you want to learn more about triangles:

brainly.com/question/2217700

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If 20 chickens laid 120<br> eggs. At this rate, how<br> many eggs would 30<br> chicken lay?

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180 eggs is the correct amount.

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3 years ago

1.) Determine the type of solutions for the function (Picture 1)

NNADVOKAT [17]

Answer:

1) 2 nonreal complex roots

2) 1 Real Solution

3) 16

4) Reflected, narrower by a factor of 2/5, slides right 4 units and slides up 6 (units)

Step-by-step explanation:

1) The graph does not intercept the x-axis, therefore, there are no real solutions at the point y = 0

We get;

y = a·x² + b·x + c

At y = 6, x = -2

Therefore;

6 = a·(-2)² - 2·b + c = 4·a - 2·b + c

6 = 4·a - 2·b + c...(1)

At y = 8, x = 0

8 = a·(0)² + b·0 + c

∴ c = 8...(2)

Similarly, we have;

At y = 8, x = -4

8 = a·(-4)² - 4·b + c = 16·a - 4·b + 8

16·a - 4·b = 0

∴ b = 16·a/4 = 4·a

b = 4·a...(3)

From equation (1), (2) and (3), we have;

6 = 4·a - 2·b + c

∴ 6 = b - 2·b + 8 = -b + 8

6 - 8 = -b

∴ -b = -2

b = 2

b = 4·a

∴ a = b/4 = 2/4 = 1/2

The equation is therefor;

y = (1/2)·x² + 2·x + 8

Solving we get;

x = (-2 ± √(2² - 4 × (1/2) × 8))/(2 × (1/2))

x =( -2 ± √(-12))/1 = -2 ± √(-12)

Therefore, we have;

2 nonreal complex roots

2) Give that the graph of the function touches the x-axis once, we have;

1 Real Solution

3) The given function is f(x) = 2·x² + 8·x + 6

The general form of the quadratic function is f(x) = a·x² + b·x + c

Comparing, we have;

a = 2, b = 8, c = 6

The discriminant of the function, D = b² - 4·a·c, therefore, for the function, we have;

D = 8² - 4 × 2 × 6 = 16

The discriminant of the function, D = 16

4.) The given function is g(x) = (-2/5)·(x - 4)² + 6

The parent function of a quadratic equation is y = x²

A vertical translation is given by the following equation;

y = f(x) + b

A horizontal to the right by 'a' translation is given by an equation of the form; y = f(x - a)

A vertical reflection is given by an equation of the form; y = -f(x) = -x²

A narrowing is given by an equation of the form; y = b·f(x), where b < 1

Therefore, the transformations of g(x) from the parent function are;

g(x) is a reflection of the parent function, with the graph of g(x) being narrower by 2/5 than the graph of the parent function. The graph of g(x) is shifted right by 4 units and is then slides up by 6 units.

7 0

2 years ago

which is the factorization of x3 8? (x 2)(x2 – 2x 4) (x – 2)(x2 2x 4) (x 2)(x2 – 2x 8) (x – 2)(x2 2x 8)

kobusy [5.1K]

Hello,
+++++++++++++++++++++
Answer B

(x^3-2^3=(x-2)(x²+2x+4)
++++++++++++++++++++++

3 0

3 years ago

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bija089 [108]

Answer:

Correct option is