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Alchen [17]
3 years ago
11

Which algebraic expression has a term with a coefficient of 9? O A. 6(x + 5) O B. 6x - 9 O C. 6+ x - 9 O D. 9x=6​

Mathematics
1 answer:
Dimas [21]3 years ago
7 0

Answer:

D is the answer to your question

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Find the area of the other polygon area=__ft
Colt1911 [192]

Answer:

108ft^2

Step-by-step explanation:

since you going up in value, you mulitple 3*9 to get 27. that means you have to multiple the other value of the bigger rectangle by 2. therefore 6*18=108

3 0
3 years ago
MN0 with vertices M(4,-5), N(5,-8), O(8, -6) ; translate the new figure M'N'O' (-2,5)
Aleks04 [339]
If you are translating a graph (-2,5), the easiest way to find the new points is to subtract the x value of each coordinate (M, N, and O) by 2. This will give you the new x values for the translated coordinates. Then, to find the new y values, add 5 to the y values of each coordinate. Then put the new x value and new y value in the form of a coordinate, and that will be your answer.
5 0
3 years ago
A table decreased in price by 3/5.After the reduction it was priced at £32. What was the original
Stels [109]

Answer:

96 not too sure but i hope it is right sry.

Step-by-step explanation:

32x3=96

3 0
2 years ago
3) m_2 = x + 70<br>60°<br>A) -10<br>C) 11<br>B)-6<br>D) 9​
kogti [31]

Answer:

C

Step-by-step explanation:

7 0
3 years ago
Which of the following sets are subspaces of R3 ?
Ratling [72]

Answer:

The following are the solution to the given points:

Step-by-step explanation:

for point A:

\to A={(x,y,z)|3x+8y-5z=2} \\\\\to  for(x_1, y_1, z_1),(x_2, y_2, z_2) \varepsilon A\\\\ a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)

                                        =3(aX_l +bX_2) + 8(ay_1 + by_2) — 5(az_1+bz_2)\\\\=a(3X_l+8y_1- 5z_1)+b (3X_2+8y_2—5z_2)\\\\=2(a+b)

The set A is not part of the subspace R^3

for point B:

\to B={(x,y,z)|-4x-9y+7z=0}\\\\\to for(x_1,y_1,z_1),(x_2, y_2, z_2) \varepsilon  B \\\\\to a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)

                                             =-4(aX_l +bX_2) -9(ay_1 + by_2) +7(az_1+bz_2)\\\\=a(-4X_l-9y_1+7z_1)+b (-4X_2-9y_2+7z_2)\\\\=0

\to a(x_1,y_1,z_1)+b(x_2, y_2, z_2) \varepsilon  B

The set B is part of the subspace R^3

for point C: \to C={(x,y,z)|x

In this, the scalar multiplication can't behold

\to for (-2,-1,2) \varepsilon  C

\to -1(-2,-1,2)= (2,1,-1) ∉ C

this inequality is not hold

The set C is not a part of the subspace R^3

for point D:

\to D={(-4,y,z)|\ y,\ z \ arbitrary \ numbers)

The scalar multiplication s is not to hold

\to for (-4, 1,2)\varepsilon  D\\\\\to  -1(-4,1,2) = (4,-1,-2) ∉ D

this is an inequality, which is not hold

The set D is not part of the subspace R^3

For point E:

\to E= {(x,0,0)}|x \ is \ arbitrary) \\\\\to for (x_1,0 ,0) ,(x_{2},0 ,0) \varepsilon E \\\\\to  a(x_1,0,0) +b(x_{2},0,0)= (ax_1+bx_2,0,0)\\

The  x_1, x_2 is the arbitrary, in which ax_1+bx_2is arbitrary  

\to a(x_1,0,0)+b(x_2,0,0) \varepsilon  E

The set E is the part of the subspace R^3

For point F:

\to F= {(-2x,-3x,-8x)}|x \ is \ arbitrary) \\\\\to for (-2x_1,-3x_1,-8x_1),(-2x_2,-3x_2,-8x_2)\varepsilon  F \\\\\to  a(-2x_1,-3x_1,-8x_1) +b(-2x_1,-3x_1,-8x_1)= (-2(ax_1+bx_2),-3(ax_1+bx_2),-8(ax_1+bx_2))

The x_1, x_2 arbitrary so, they have ax_1+bx_2 as the arbitrary \to a(-2x_1,-3x_1,-8x_1)+b(-2x_2,-3x_2,-8x_2) \varepsilon F

The set F is the subspace of R^3

5 0
3 years ago
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