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san4es73 [151]
3 years ago
7

Does this table represent a linear equation?

Mathematics
1 answer:
olga2289 [7]3 years ago
5 0
No it does not
Hope this helps
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Find the area of each sector.
sergey [27]

Answer:

D. \frac{525\pi}{8} m²

Step-by-step explanation:

Area of the sector = ½*r²∅

Where,

Radius (r) = 15 m

Angle in radians (∅) = \frac{7\pi}{12}

Plug in the values into the equation

Area of sector = \frac{1}{2} * 15^2 * \frac{7\pi}{12}

Area of sector = \frac{1}{2} * 225 * \frac{7\pi}{12}

Area of sector = \frac{1*225*7\pi}{2*12}

Area of sector = \frac{1,575\pi}{24}

Simplify

Area of sector = \frac{525\pi}{8}

5 0
3 years ago
Help pleaseeeeeeeeeeeeee!!!!!!!!!!!!!!!! T^T
artcher [175]
1 + (-5/8) = 3/8

Explanation: If the coffee wasn’t drank it would’ve been at fraction 1 (or 100%)

Since he drank 5/8 of it. The coffee left is now 1-(5/8)

But since the question has asked to give the equation in addition format. Put a plus sign in between the two.

Therefore,
1 + (-5/8)

Solve it,
= 1-(5/8)
= (8-5)/8
= 3/8

Hence the equation is:

1 + (-5/8) = 3/8.
8 0
3 years ago
Read 2 more answers
Which measure is greater?? One inch or one centimeter
Nimfa-mama [501]

One inch is the greater measure

8 0
3 years ago
5{-3.4k-7} + {3k+21}
nekit [7.7K]
First step is to do distributive property.
 
so you multiply 5 by -3.4k and -7 

your equation now is -17k-35+(3k+21)
 
Then you add like terms 

add -17k+3k to give you -14k-35+21 
 
then you add like terms again

add -35+21 to give you -14

so now your equation is -14k-14 

So that is your answer simplified 
6 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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