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lara [203]
3 years ago
11

Find the area of the shape with the radius of 10

Mathematics
1 answer:
lukranit [14]3 years ago
3 0

Answer: 314.16

Step-by-step explanation: To find the area of a circle you have to square the radius then multiply by pi. 10 squared is 100 then multiplied by pi is 314.16 (rounded to the nearest hundredth).

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How are these triangles congruent??????
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Answer:

SAS Test

Step-by-step explanation:

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I desperately need help !
Lorico [155]

Answer:

t = 9.57

Step-by-step explanation:

We can use trig functions to solve for the t

Recall the 3 main trig ratios

Sin = opposite / hypotenuse

Cos = adjacent / hypotenuse

Tan = opposite / adjacent.

( note hypotenuse = longest side , opposite = side opposite of angle and adjacent = other side )

We are given an angle as well as its opposite side length ( which has a measure of 18 ) and we need to find its adjacent "t"

When dealing with the opposite and adjacent we use trig ratio tan.

Tan = opp / adj

angle measure = 62 , opposite side length = 18 and adjacent = t

Tan(62) = 18/t

we now solve for t

Tan(62) = 18/t

multiply both sides by t

Tan(62)t = 18

divide both sides by tan(62)

t = 18/tan(62)

t = 9.57

And we are done!

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Oduvanchick [21]

Answer:

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Step-by-step explanation:

A) We can prove it as follows:

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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||.

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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.  

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