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Strike441 [17]
3 years ago
7

If (9.1) and (1.9) are the endpoints of a diameter of a circle, what is the equation of the cire

Mathematics
1 answer:
likoan [24]3 years ago
6 0
The answer will be B
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Llll answer this an i will mark u brainliest​
damaskus [11]
2 / -18 and 4/ a (1/2 ) I THINKKK
3 0
2 years ago
Can you please help me with this problem?
Hoochie [10]
Here you have to find which each variable is, for this you start of picking one equation,

x + 2y + 6z = 4
-3x + 2y - 2 = -4
4x + 2z = 16

depending the equation you pick you multiply that by a certain number that will give you the opposite of one of the other equations,
-1(x + 2y + 6z = 4)
= -x -2y - 6z = -4

With this you add or subtract it with the equation that has the same number or variable, or both,
In this case it will be the equation,
-3x + 2y + 6z = 4
You can use this one or the third equation since both have a positive 2y which will cancel with -2y from the new equation,
-x - 2y - 6z = -4
-3x + 2y -z = -4

= -4x -7z = -8
Now you since you just eliminated the variable (y) you now have 2 variables, and the last equation has only 2 variables, meaning now you find the answer to those to equations,

-4x -7z = -8
4x + 2z = 16

= -5z = 8
Now leave the variable by itself,
z = 8/5
Now you found the variable (z), with this just substitute on one of the equations we used to find (z) so you can find (x), after that substitute those answered to on of the original equations so you can find (y)




6 0
3 years ago
PLSSSSS HELP ME, Write a formula for the sum of the first three terms, a1+a2+a3, Your answer should only include variables a1 an
Gwar [14]
S3= 3/2 [2a1+(3-1)d]
6 0
3 years ago
How do I do this problem I am stuck .
marta [7]
1) Add 9 to both sides. On the left side of the equal sign it will cancel out. on the right side of the equal sign because it is now a different sign equation you subtract -193 from 9 which equals -184
2) now you are left with to= -184
to get the variable by itself you need to divide both sides by 4. on the left side of the equal sign you have p by itself now , on the right you divide -184 by 4 and your answer comes out to be p= -46
3) p= -46 is your final answer!
7 0
3 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
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