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julia-pushkina [17]
3 years ago
15

Use addition to solve the linear system of equations. Include all of your work in your final answer.

Mathematics
1 answer:
nadezda [96]3 years ago
3 0
You just have to add one eqaution to another :
-x + x -y + 2y = 7 + 7
y = 14
Then apply y = 14 to one of the equations :
eg: x + 2(14) = 7
x + 28 = 7
x = - 21
The awnser is x= -21 y= 14
I hope i awnsered your question
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As used in line 53, "dominion" most nearly means
drek231 [11]

Choice B - Supremacy

As used in line 53, "dominion" most nearly means supremacy.

In the first paragraph of Passage 2, when discussing changing social relations, Mill writes that in her time there was "a just equality, instead of the supremacy of the strongest." In this context of a society where some once had much more power than others, the word "dominion" almost means superiority or greater power.

Choices A, C, and D are incorrect because, in the context of a paragraph discussing differences in the amount of power members of a society have, "dominion" means dominion or greater power, not omnipotence or the state of being all-powerful (choice A), possession ( choice C) or territory (choice D).

For more information about dominion, visit

brainly.com/question/11588378

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6 0
1 year ago
What additional information could be used to prove that ΔXYZ ≅ ΔFEG using ASA or AAS? Check all that apply.
Maksim231197 [3]

Answer: ∠Z ≅ ∠G and XZ ≅ FG or ∠Z ≅ ∠G and XY ≅ FE  are the additional information could be used to prove that ΔXYZ ≅ ΔFEG using ASA or AAS.

Step-by-step explanation:

Given: ΔXYZ and ΔEFG such that ∠X=∠F

To prove they are congruent by using ASA or AAS conruency criteria

we need only one angle and side.

1. ∠Z ≅ ∠G(angle) and XZ ≅ FG(side)

so we can apply ASA  such that ΔXYZ ≅ ΔFEG.

2. ∠Z ≅ ∠G (angle)and ∠Y ≅ ∠E (angle), we need one side which is not present here.∴we can not apply ASA  such that ΔXYZ ≅ ΔFEG.

3. XZ ≅ FG (side) and ZY ≅ GE (side), we need one angle which is not present here.∴we can not apply ASA  such that ΔXYZ ≅ ΔFEG.

4. XY ≅ EF(side) and ZY ≅ FG(side), not possible.

5. ∠Z ≅ ∠G(angle) and XY ≅ FE(side),so we can apply ASA  such that

ΔXYZ ≅ ΔFEG.

4 0
3 years ago
Read 2 more answers
What graph represents number 17? (y - in; tercept = 1; t=1;slope= 1 2)
Fittoniya [83]

Answer:

The answer to this question is graph B

8 0
2 years ago
Let V denote the set of ordered triples (x, y, z) and define addition in V as in
icang [17]

Answer:

a) No

b) No

c) No

d) No

Step-by-step explanation:

Remember, a set V wit the operations addition and scalar product is a vector space if the following conditions are valid for all u, v, w∈V and for all scalars c and d:

1. u+v∈V

2. u+v=v+u

3. (u+v)+w=u+(v+w).

4. Exist 0∈V such that u+0=u

5. For each u∈V exist −u∈V such that u+(−u)=0.

6. if c is an escalar and u∈V, then cu∈V

7. c(u+v)=cu+cv

8. (c+d)u=cu+du

9. c(du)=(cd)u

10. 1u=u

let's check each of the properties for the respective operations:

Let u=(u_1,u_2,u_3), v=(v_1,v_2,v_3)

Observe that  

1. u+v∈V

2. u+v=v+u, because the adittion of reals is conmutative

3. (u+v)+w=u+(v+w). because the adittion of reals is associative

4. (u_1,u_2,u_3)+(0,0,0)=(u_1+0,u_2+0,u_3+0)=(u_1,u_2,u_3)

5. (u_1,u_2,u_3)+(-u_1,-u_2,-u_3)=(0,0,0)

then regardless of the escalar product, the first five properties are met for a), b), c) and d). Now let's verify that properties 6-10 are met.

a)

6. c(u_1,u_2,u_3)=(cu_1,u_2,cu_3)\in V

7.

c(u+v)=c(u_1+v_1,u_2+v_2,u_3+v_3)=(c(u_1+v_1),u_2+v_2,c(u_3+v_3))\\=(cu_1+cv_1,u_2+v_2,cu_3+cv_3)=c(u_1,u_2,u_3)+c(v_1,v_2,v_3)=cu+cv

8.

(c+d)u=(c+d)(u_1,u_2,u_3)=((c+d)u_1,u_2,(c+d)u_3)=\\=(cu_1+du_1,u_2,cu_3+du_3)\neq (cu_1+du_1,2u_2,cu_3+du_3)=cu+du

Since 8 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product a(x,y,z)=(ax,y,az)

b)  6. c(u_1,u_2,u_3)=(cu_1,0,cu_3)\in V

7.

c(u+v)=c(u_1+v_1,u_2+v_2,u_3+v_3)=(c(u_1+v_1),0,c(u_3+v_3))\\=(cu_1+cv_1,0,cu_3+cv_3)=c(u_1,u_2,u_3)+c(v_1,v_2,v_3)=cu+cv

8.

(c+d)u=(c+d)(u_1,u_2,u_3)=((c+d)u_1,0,(c+d)u_3)=\\=(cu_1+du_1,0,cu_3+du_3)=(cu_1,0,cu_3)+(du_1,0,du_3) =cu+du

9.

c(du)=c(d(u_,u_2,u_3))=c(du_1,0,du_3)=(cdu_1,0,cdu_3)=(cd)u

10

1u=1(u_1,u_2,u3)=(1u_1,0,1u_3)=(u_1,0,u_3)\neq(u_1,u_2,u_3)

Since 10 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product a(x,y,z)=(ax,0,az)

c) Observe that 1u=1(u_1,u_2,u3)=(0,0,0)\neq(u_1,u_2,u_3)

Since 10 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product a(x,y,z)=(0,0,0).

d)  Observe that 1u=1(u_1,u_2,u3)=(2*1u_1,2*1u_2,2*1u_3)=(2u_1,2u_2,2u_3)\neq(u_1,u_2,u_3)=u

Since 10 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product a(x,y,z)=(2ax,2ay,2az).

8 0
3 years ago
The mean weight of 25 student in a class is 58kg. The mean weight of a class of 29 students is 62kg. Find the mean weight of all
WARRIOR [948]

Answer:

weight of 25 students

25 × 58 = 1450kg

weight of 29 students

29 × 62 = 1798kg

total students = 25 + 29 = 54

total weight of students = 1450 + 1798 = 3248

mean weight of all students = 3248/54 = 60.19kg

6 0
2 years ago
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