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3 years ago

Two positive integers have their Hcf as 12 and their product as 6336. The number of possible pairs for the numbers is

Mathematics

2 answers:

oksano4ka [1.4K]3 years ago

7 0

the answer is 4? I am not sure

melamori03 [73]3 years ago

7 0

Answer:

The number of possible pairs for the number (B) 3

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What fractions are equivalent to these fractions<br> 4/5=<br> 3/8=<br> 1/6=

lukranit [14]

Answer:

1 and the 3rd

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

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 $ax_1+bx_2$ is 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

2 years ago

Write sin 29 degrees 32 minutes in terms of it cofunction

12345 [234]

Answer:

Sin 29° 32' = Cos 60° 28'

Step-by-step explanation:

Here in this problem, we have to write Sin 29° 32' in terms of its co-function.

We know that co-function of Sin Ф = Cos ( 90° - Ф ).

Therefore, we have to find a complementary angle of 29° 32'.

So, ( 90° - 29° 32' ) = 60° 28'

Therefore, Sin 29° 32' = Cos 60° 28' ( Answer )

5 0

2 years ago

the 41st term of the arithmetic sequence is -539 and the 54th term is -708. Find the value of the 80th term

yulyashka [42]

Answer:

$a _{n} = a _{1} + (n - 1)d \\ a _{41} = - 534 \\ a _{54} = - 708 \\ a _{41} = a _{1} + (41 - 1)d = - 534 \\ a _{1} + 40d = - 534 \: \: ...(1)\\ a _{54} = a _{1} + (54 - 1)d =- 708 \\ a _{1} + 53d = - 708 \: \: ...(2)$

Try to solve the similtanious equation to find the first term and the common difference, then use that information to figure out the 80th term... Update me if you need further help ...

7 0

2 years ago

(LC)

serg [7]

10

1240 x 10 = 12400
12400 = 12400

8 0

2 years ago