Select all the ratios that are equivalent to each other.

Mathematics

1 answer:

lora16 [44]3 years ago

7 0

Well, all I know is that 4:7 and 16:28 are equivalent. I hope that helps

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Solve the triangle A = 2 B = 9 C =8

VARVARA [1.3K]

Answer:

$\begin{gathered} A=\text{ 12}\degree \\ B=\text{ 114}\degree \\ C=54\degree \end{gathered}$

Step-by-step explanation:

To calculate the angles of the given triangle, we can use the law of cosines:

$\begin{gathered} \cos (C)=\frac{a^2+b^2-c^2}{2ab} \\ \cos (A)=\frac{b^2+c^2-a^2}{2bc} \\ \cos (B)=\frac{c^2+a^2-b^2}{2ca} \end{gathered}$

Then, given the sides a=2, b=9, and c=8.

$\begin{gathered} \cos (A)=\frac{9^2+8^2-2^2}{2\cdot9\cdot8} \\ \cos (A)=\frac{141}{144} \\ A=\cos ^{-1}(\frac{141}{144}) \\ A=11.7 \\ \text{ Rounding to the nearest degree:} \\ A=12º \end{gathered}$

For B:

$\begin{gathered} \cos (B)=\frac{8^2+2^2-9^2}{2\cdot8\cdot2} \\ \cos (B)=\frac{13}{32} \\ B=\cos ^{-1}(\frac{13}{32}) \\ B=113.9\degree \\ \text{Rounding:} \\ B=114\degree \end{gathered}$ $\begin{gathered} \cos (C)=\frac{2^2+9^2-8^2}{2\cdot2\cdot9} \\ \cos (C)=\frac{21}{36} \\ C=\cos ^{-1}(\frac{21}{36}) \\ C=54.3 \\ \text{Rounding:} \\ C=\text{ 54}\degree \end{gathered}$

3 0

1 year ago

All vectors are in Rn. Check the true statements below:

Oduvanchick [21]

Answer:

A), B) and D) are true

Step-by-step explanation:

A) We can prove it as follows:

$Proy_{cv}y=\frac{(y\cdot cv)}{||cv||^2}cv=\frac{c(y\cdot v)}{c^2||v||^2}cv=\frac{(y\cdot v)}{||v||^2}v=Proy_{v}y$

B) When you compute the product Ax, the i-th component is the matrix of the i-th column of A with x, denote this by Ai x. Then, we have that $||Ax||=\sqrt{(A_1 x)^2+\cdots (A_n x)^2}$ . Now, the colums of A are orthonormal so we have that (Ai x)^2=x_i^2. Then $||Ax||=\sqrt{(x_1)^2+\cdots (x_n)^2}=||x||$ .

C) Consider $S=\{(0,2),(2,0)\}\subseteq \mathbb{R}^2$ . This set is orthogonal because $(0,2)\cdot(2,0)=0(2)+2(0)=0$ , but S is not orthonormal because the norm of (0,2) is 2≠1.

D) Let A be an orthogonal matrix in $\mathbb{R}^n$ . Then the columns of A form an orthonormal set. We have that $A^{-1}=A^t$ . To see this, note than the component $b_{ij}$ of the product $A^t A$ is the dot product of the i-th row of $A^t$ and the jth row of $A$ . But the i-th row of $A^t$ is equal to the i-th column of $A$ . If i≠j, this product is equal to 0 (orthogonality) and if i=j this product is equal to 1 (the columns are unit vectors), then $A^t A=I$

E) Consider S={e_1,0}. S is orthogonal but is not linearly independent, because 0∈S.

In fact, every orthogonal set in R^n without zero vectors is linearly independent. Take a orthogonal set $\{u_1,u_2\cdots u_p\}$ and suppose that there are coefficients a_i such that $a_1u_1+a_2u_2\cdots a_nu_n=0$ . For any i, take the dot product with u_i in both sides of the equation. All product are zero except u_i·u_i=||u_i||. Then $a_i||u_i||=0$ then $a_i=0$ .