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Llana [10]
3 years ago
9

Similar triangles vs con congruent explanation

Mathematics
1 answer:
Licemer1 [7]3 years ago
4 0

Answer:

Similar triangle means that only the angles are the same, but the sides are not. Congruent means the angles and sides are the same.

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eimsori [14]
Wouldn't all the angles equal 180? We've been trying to learn these but I think so. Subtract both numbers by 180 and it should be the answer. Sorry if it's wrong but I'm not sure. I kinda hope this helps you any
5 0
3 years ago
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3. Find the lengths of d<br> d<br> 15<br> 7<br> 6<br> 10
siniylev [52]

Answer:

15

Step-by-step explanation:

As you can see on the side it says 15 and that is the length of the structure and it would be the same all around

4 0
3 years ago
For questions 3 - 4, identify the symbol that correctly relates the two numbers, and then determine
natka813 [3]

Answer:

Goodluck

Step-by-step explanation:

6 0
3 years ago
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The vertex of the parabola below is at the point (2,4), and the point (3,6) is
lukranit [14]

Answer:

A. y= 2(x - 2)2 + 4

Step-by-step explanation:

vertex having an x value of 2 means the phrase in paren must be x - 2.

This eliminates B and D

vertex having a y value of 4 is of no help as the remaining two equations will result in 4 if the x term is 2

plugging in x = 3 makes y = 6 in equation A, but not in equation C where y = 10

3 0
3 years ago
Please determine whether the set S = x^2 + 3x + 1, 2x^2 + x - 1, 4.c is a basis for P2. Please explain and show all work. It is
ohaa [14]

The vectors in S form a basis of P_2 if they are mutually linearly independent and span P_2.

To check for independence, we can compute the Wronskian determinant:

\begin{vmatrix}x^2+3x+1&2x^2+x-1&4\\2x+3&4x+1&0\\2&4&0\end{vmatrix}=4\begin{vmatrix}2x+3&4x+1\\2&4\end{vmatrix}=40\neq0

The determinant is non-zero, so the vectors are indeed independent.

To check if they span P_2, you need to show that any vector in P_2 can be expressed as a linear combination of the vectors in S. We can write an arbitrary vector in P_2 as

p=ax^2+bx+c

Then we need to show that there is always some choice of scalars k_1,k_2,k_3 such that

k_1(x^2+3x+1)+k_2(2x^2+x-1)+k_34=p

This is equivalent to solving

(k_1+2k_2)x^2+(3k_1+k_2)x+(k_1-k_2+4k_3)=ax^2+bx+c

or the system (in matrix form)

\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}a\\b\\c\end{bmatrix}

This has a solution if the coefficient matrix on the left is invertible. It is, because

\begin{vmatrix}1&1&0\\3&1&0\\1&-1&4\end{vmatrix}=4\begin{vmatrix}1&2\\3&1\end{vmatrix}=-20\neq0

(that is, the coefficient matrix is not singular, so an inverse exists)

Compute the inverse any way you like; you should get

\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}^{-1}=-\dfrac1{20}\begin{bmatrix}4&-8&0\\-12&4&0\\-4&3&-5\end{bmatrix}

Then

\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}^{-1}\begin{bmatrix}a\\b\\c\end{bmatrix}

\implies k_1=\dfrac{2b-a}5,k_2=\dfrac{3a-b}5,k_3=\dfrac{4a-3b+5c}{20}

A solution exists for any choice of a,b,c, so the vectors in S indeed span P_2.

The vectors in S are independent and span P_2, so S forms a basis of P_2.

5 0
3 years ago
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