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marshall27 [118]
3 years ago
15

If the angle of elevation to the top of a tree is 18° and the tree's height is 40 feet, how far are you away from the tree?

Mathematics
1 answer:
Darina [25.2K]3 years ago
3 0
Check the picture below.

make sure your calculator is in Degree mode.

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What is parallel to y=-2x+5(1,2)(1,-2)(-2,-7)(2,5)
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1) A parallel line has the same slope as this one y=-2x+5 whose slope is -2

So to find out which parallel line to y=-2x+5contains

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Rectangle is 5 feet wide and 11 feet long. What is the perimeter
Ahat [919]

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Just add all side

5+11+5+11=32 feet

4 0
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I need help!!!! Can anyone answer as fast as they can???
Morgarella [4.7K]
X= 180
now you have to multiply all them with 180
5(180)= 900
2(180)= 360


8 0
3 years 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.  

5 0
3 years ago
11 3/4-7 1/4 evaluate answer need asap
Debora [2.8K]

Answer:

−6

Step-by-step explanation:

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