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DaniilM [7]
3 years ago
13

Describe lengths of three segments that could not be used to form a triangle. Segments with lengths of 5 in., 5 in., and ______

in. Cannot form a triangle.
Mathematics
1 answer:
Vesna [10]3 years ago
8 0

Answer:

c = 11

Step-by-step explanation:

Given

Let the three sides be a, b and c

Such that:

a = b= 5

Required

Find c such that a, b and c do not form a triangle

To do this, we make use of the following triangle inequality theorem

a + b > c

a + c > b

b + c > a

To get a valid triangle, the above inequalities must be true.

To get an invalid triangle, at least one must not be true.

Substitute: a = b= 5

5 + 5 > c ===> 10 > c

5 + c > 5  ===> c > 5 - 5 ===> c > 0

5 + c > 5  ===> c > 5 - 5 ===> c > 0

The results of the inequality is: 10 > c and c > 0

Rewrite as: c > 0 and c < 10

0 < c < 10

This means that, the values of c that make a valid triangle are 1 to 9 (inclusive)

Any value outside this range, cannot form a triangle

So, we can say:

c = 11, since no options are given

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Answer:

The components of \vec{b} parallel and perpendicular to \vec {a} are \vec {b}_{\parallel} = \frac{3}{2}\,i-\frac{1}{2}\,j and \vec b _{\perp} = \frac{1}{2}\,i+\frac{3}{2}\,j-3\,k, respectively.

Step-by-step explanation:

Let be \vec b = 2\,i+j-3\,k and \vec a = 3\,i-j, the component of \vec b parallel to \vec a is calculated by the following expression:

\vec b_{\parallel} = (\vec b \bullet \hat{a}) \cdot \hat{a}

Where \hat{a} is the unit vector of \vec a, dimensionless and \bullet is the operator of scalar product.

The unit vector of \vec a is:

\hat{a} = \frac{\vec {a}}{\|\vec a\|}

Where \|\vec {a}\| is the norm of \vec a, whose value is determined by Pythagorean Theorem.

The component of \vec{b} parallel to \vec {a} is:

\|\vec {a}\| = \sqrt{3^{2}+(-1)^{2}+0^{2}}

\|\vec {a}\| = \sqrt{10}

\hat{a} = \frac{1}{\sqrt{10}} \cdot (3\,i-j)

\hat{a} = \frac{3}{\sqrt{10}}\,i -\frac{1}{\sqrt{10}} \,j

\vec{b}\bullet \hat{a} = (2)\cdot \left(\frac{3}{\sqrt{10}} \right)+(1)\cdot \left(-\frac{1}{\sqrt{10}} \right)+(-3)\cdot \left(0\right)

\vec b \bullet \hat{a} = \frac{5}{\sqrt{10}}

\vec b_{\parallel} = \frac{5}{\sqrt{10}}\cdot \left(\frac{3}{\sqrt{10}}\,i-\frac{1}{\sqrt{10}}\,j  \right)

\vec {b}_{\parallel} = \frac{3}{2}\,i-\frac{1}{2}\,j

Now, the component of \vec {b} perpendicular to \vec{a} is found by vector subtraction:

\vec{b}_{\perp} = \vec {b}-\vec {b}_{\parallel}

If \vec b = 2\,i+j-3\,k and \vec {b}_{\parallel} = \frac{3}{2}\,i-\frac{1}{2}\,j, then:

\vec{b}_{\perp} = (2\,i+j-3\,k)-\left(\frac{3}{2}\,i-\frac{1}{2}\,j  \right)

\vec b _{\perp} = \frac{1}{2}\,i+\frac{3}{2}\,j-3\,k

4 0
3 years ago
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