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gizmo_the_mogwai [7]
4 years ago
7

Look at the following graph of the given equation. Determine whether the equation is the function. Explain why or why not.

Mathematics
1 answer:
Drupady [299]4 years ago
3 0

Answer:

Step-by-step explanation:

It is a function. Every x has only 1 y associated with it. You cannot find an exception to this rule.

Stated another more common  way: if you take a ruler and lay it flat on the graph so that the ruler is parallel to the y axis, you will find the ruler will go through only one y for every x. That's called the vertical line test.

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Answer:

800

Step-by-step explanation:

Hope it helps you out

-sammy

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3 years ago
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Determine whether the triangles are similar. or not.
wariber [46]
I believe the correct answer C
4 0
3 years ago
Find the volume of a right circular cone that has a height of 3.5 ft and a base with a radius of 18.9 ft. Round your answer to t
11111nata11111 [884]

Answer: 1309.24 I used calculations and I think this is right

Step-by-step explanation:

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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
4 years ago
Anyone can help ? I have no clue what the answer maybe
WINSTONCH [101]
Since both the triangles are similar,

(y-2)/(y+4) = 4/8

8(y-2) = 4(y+4)

8y - 16 = 4y + 16

8y - 4y = 16 + 16

4y = 32

y = 32/4

y = 8


The perimeter of triangle PQR:
4 + 6 + (8-2)

= 10 + 6

= 16 ft.
5 0
3 years ago
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