How many triangles can be constructed with sides measuring 15 cm, 7 cm, and 5 cm? none more than one one

Mathematics

2 answers:

Alecsey [184]3 years ago

6 0

Answer:

it is none

Step-by-step explanation:

i got a 80 becasue its not one its none just took the test

IrinaK [193]3 years ago

4 0

<h3>Answer: one</h3>

Reason:

The SSS (side side side) congruence rule says that two triangles are congruent if we have three pairs of sides that are the same length. Any triangle you construct that has side lengths 15, 7, and 5 will be congruent to the original triangle by the SSS rule. You may have to translate, rotate or reflect the second triangle so that it lines up perfectly with the first triangle.

Alternatively, consider triangle ABC that is the given triangle. Let AB = 15, BC = 7, and AC = 5. Then consider another triangle DEF such that DE = 15, EF = 7, and DF = 5. It is possible to do a series of translations, rotations, or reflections such that we get triangle DEF to overlap triangle ABC perfectly.

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Find a fundamental matrix solution for the system x 0 1 7x1 + 4x2 + 12x3, x 0 2 x1 + 2x2 + x3, x 0 3 −3x1 − 2x2 − 5x3. Then find

liq [111]

Here is the correct format for the question.

$x'_1 = 7x_1 +4x_2+ 12x_3 , \ \ x'_2 = x_1 + 2x_2 + x_3 , \ \ x'_3 = -3x_1 -2x_2 -5x_3$ . Then find the solution that satisfies $x \limits ^{\to} = \left[\begin{array}{c}0\\1\\-2\end{array}\right]$

Answer:

$\mathbf{c_1 = -3, c_2 = 2, c_3 = 2}$

Step-by-step explanation:

From the figures given above:

the matrix can be computed as,

$\left[\begin{array}{ccc}7&4&12\\1&2&1\\-3&-2&-5\end{array}\right] x \limits ^{\to} = \left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right] = x \limits ^{\to} = \left[\begin{array}{c}x_1'\\x_2'\\x_3'\end{array}\right]$

The first thing we need to carry out is to determine the eigenvalues of A,

where:

$A = \left[\begin{array}{ccc}7&4&12\\1&2&1\\-3&-2&-5\end{array}\right]$

$|A-rI|=0$

$\begin {vmatrix} 7-r&4&12\\1&2-r&1\\-3&-2&-5-r \end {vmatrix}=0$

the eigenvalues are r = 0, 1, 3

However, the eigenvector correlated to each eigenvalue can be calculated as follows.

suppose r = 0

(A - rI) x = 0

$\left[\begin{array}{ccc}7&4&12\\1&2&1\\-3&-2&-5\end{array}\right] \left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right] = \left[\begin{array}{c}0\\0\\0\end{array}\right]$

now the eigenvector is $\left[\begin{array}{c}-4\\1\\2\end{array}\right]$

However, for eigenvalue = 1, we have : $\left[\begin{array}{c}-4\\1\\2\end{array}\right]$

for eigenvalue = 3, we have: $\left[\begin{array}{c}-2\\-1\\1\end{array}\right]$

The solution now can be computed as :

$x(t)= c_1 \left[\begin{array}{c}-4\\1\\2\end{array}\right] + c_2e^t \left[\begin{array}{c}-4\\3\\1\end{array}\right]+ c_3e^{3t} \left[\begin{array}{c}-2\\-1\\1\end{array}\right]$