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valentinak56 [21]
3 years ago
14

Find the factors of each number in classify each number as a prime or composite 18,37,48

Mathematics
1 answer:
zepelin [54]3 years ago
7 0

18=1,2,3,9,18

37=1,37

48=1,2,3,16,24,12,4,6,8,48

Prime no. is a number that is divisible only by itself and 1

So, here 37 is a prime no.

18 and 48 are composite no.s

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Usimov [2.4K]

0,3/4

1/2,1


X being the first, y second term verify the equation

4 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
Which values for h and k are used to write the function f of x = x squared 12 x 6 in vertex form?
Nesterboy [21]

The values of h and k when f(x) = x^2 + 12x + 6 is in vertex form is -6 and -30

<h3>How to rewrite in vertex form?</h3>

The equation is given as:

f(x) = x^2 + 12x + 6

Rewrite as:

x^2 + 12x + 6 = 0

Subtract 6 from both sides

x^2 + 12x = -6

Take the coefficient of x

k = 12

Divide by 2

k/2 = 6

Square both sides

(k/2)^2 = 36

Add 36 to both sides of x^2 + 12x = -6

x^2 + 12x + 36= -6 + 36

Evaluate the sum

x^2 + 12x + 36= 30

Express as perfect square

(x + 6)^2 = 30

Subtract 30 from both sides

(x + 6)^2 -30 = 0

So, the equation f(x) = x^2 + 12x + 6 becomes

f(x) = (x + 6)^2 -30

A quadratic equation in vertex form is represented as:

f(x) = a(x - h)^2 + k

Where:

Vertex = (h,k)

By comparison, we have:

(h,k) = (-6,-30)

Hence, the values of h and k when f(x) = x^2 + 12x + 6 is in vertex form is -6 and -30

Read more about quadratic functions at:

brainly.com/question/1214333

#SPJ1

3 0
2 years ago
Which equation can be used to solve for x in the following diagram?
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Answer:

b

Step-by-step explanation:

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