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Dafna1 [17]
2 years ago
12

Which statement best describes the area of Triangle ABC shown below? A triangle ABC is shown on a grid. The vertex A is on order

ed pair 3 and 5, vertex B is on ordered pair 4 and 1, and the vertex C is on ordered pair 2 and 1. It is twice the area of a square of side length 2 units. It is one-half the area of a square of side length 2 units. It is twice the area of a rectangle of length 2 units and width 4 units.
Mathematics
1 answer:
Alinara [238K]2 years ago
3 0

Answer:

( x , y ) = ( -1 , - 5 )

Step-by-step explanation:

This was the answer when i did it.

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Help me plz this is my last question of the day but plz help me
gizmo_the_mogwai [7]

Answer:

Additive Inverse Property.

Step-by-step explanation:

the value that brings you back to the identity element under addition

So a+-a=0

Hope this helps!!!

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Plot the image of point D under a dilation about the origin (0,0) with the scale factor of 2.
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3 years ago
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CAN SOMEONE HELP ME WITH THIS MATH QUESTION ITS ABOUT TRANSLATION
earnstyle [38]

Answer:

(- 4, 1 )

Step-by-step explanation:

A translation (x + 1), y - 3 ) means add 1 to the original x- coordinate and subtract 3 from the original y- coordinate.

B = (- 5, 4 ), thus

B' = (- 5 + 1, 4 - 3 ) = (- 4, 1 )

5 0
3 years ago
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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
3 years ago
Beaker A contains 1 liter which is 30 percent oil and the rest is vinegar, thoroughly mixed up. Beaker B contains 2 liters which
alekssr [168]

Answer:

1.25 liters of oil

Step-by-step explanation:

Volume in Beaker A = 1 L

Volume of Oil in Beaker A = 1*0.3 = 0.3 L

Volume of Vinegar in Beaker A = 1*0.7 = 0.7 L

Volume in Beaker B = 2 L

Volume of Oil in Beaker B = 2*0.4 = 0.8 L

Volume of Vinegar in Beaker B = 1*0.6 = 1.2 L

If half of the contents of B are poured into A and assuming a homogeneous mixture, the new volumes of oil (Voa) and vinegar (Vva) in beaker A are:

V_{oa} = 0.3+\frac{0.8}{2} \\V_{oa} = 0.7 \\V_{va} = 0.7+\frac{1.2}{2} \\V_{va} = 1.3

The amount of oil needed to be added to beaker A in order to produce a mixture which is 60 percent oil (Vomix) is given by:

0.6*V_{total} = V_{oa} +V_{omix}\\0.6*(V_{va}+V_{oa} +V_{omix}) = V_{oa} +V_{omix}\\0.6*(1.3+0.7+V_{omix})=0.7+V_{omix}\\V_{omix}=\frac{0.5}{0.4} \\V_{omix}=1.25 \ L

1.25 liters of oil are needed.

6 0
3 years ago
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