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3 years ago

What are the three Pythagorean identities?

1 answer:

4 0

$Pythagorean \ Identity : \\\\1) \\\\sin^2\alpha +cos^2\alpha =1\\\\2)\\\\tan^2\alpha +1=sec^2\alpha \\\\ 3)\\\\1+cot^2\alpha =csc^2\alpha$

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Find the measure of

jok3333 [9.3K]

Answer:

119

Step-by-step explanation:

because if u see <EAC AND<BAD. are vertical so if u add 62°to 57° or adds up to 119

3 0

3 years ago

Jerry has 61 percent on khan. She needs to get 100 percent in one week. How much percent should she get each night to reach her

vova2212 [387]

Answer:

She would need to complete 5.57 each noght to reach her goal of 100%.

Step-by-step explanation:

39/7 is equal to 5.57

7 0

3 years ago

What is the length of segment AB?<br> Consider the diagram.<br> 7,9,18,25

Mrrafil [7]

Answer:

Step-by-step explanation:

It is congruent to the other side.

5 0

3 years ago

Can you answer these please many thanks

givi [52]

Given: (a) $2(x+3)=2x+6$ and (b) $4(y-3)=4y-12$ .

To find: The expanded form of the given expressions.

(d) $3(5-q)=15-3q$ (using $a(b-c)=ab-ac$ )

(e) $5(2c+1)=10c+5$ (using $a(b+c)=ab+ac$ )

(f) $3(2x-5)=6x-15$ (using $a(b-c)=ab-ac$ )

(g) $7(4b-1)=28b-7$ (using $a(b-c)=ab-ac$ )

(h) $3(2x+y-5)=3((2x+y)-5)$

⇒ $3(2x+y-5)=3(2x+y)-15$ (using $a(b-c)=ab-ac$ )

⇒ $3(2x+y-5)=6x+3y-15$ (using $a(b+c)=ab+ac$ )

(i) $2(6a-4b+3)=2((6a-4b)+3)$

⇒ $2(6a-4b+3)=2(6a-4b)+6$ (using $a(b+c)=ab+ac$ )

⇒ $2(6a-4b+3)=12a-8b+6$ (using $a(b-c)=ab-ac$ )

(j) $6(m+n+p)=6((m+n)+p)$

⇒ $6(m+n+p)=6(m+n)+6p$ (using $a(b+c)=ab+ac$ )

⇒ $6(m+n+p)=6m+6n+6p$ (using $a(b+c)=ab+ac$ )

(k) $y(y+2)=y^2+2y$ (using $a(b+c)=ab+ac$ )

(l) $g(g-3)=g^2-3g$ (using $a(b-c)=ab-ac$ )

(m) $n(4-n)=4n-n^2$ (using $a(b-c)=ab-ac$ )

(n) $a(b+c)=ab+ac$ (using $a(b+c)=ab+ac$ )

(o) $s(3s-4)=3s^2-4s$ (using $a(b-c)=ab-ac$ )

(p) $2x(x+5)=2x^2+10x$ (using $a(b+c)=ab+ac$ )

(q) $4y(x-3)=4xy-12y$ (using $a(b-c)=ab-ac$ )

(r) $5a(2b-5)=10ab-25a$ (using $a(b-c)=ab-ac$ )

(s) $4a(3b+2c)=12ab+8ac$ (using $a(b+c)=ab+ac$ )

(t) $5p(4p-5q)=20p^2-25pq$ (using $a(b-c)=ab-ac$ )

6 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