All categories

1 year ago

Two numbers have these properties.

Mathematics

1 answer:

taurus [48]1 year ago

8 0

With an HCF of 6 and an LCM of 60, the numbers 12 and 30 are greater than 6.

<h3>What are mathematical operations?</h3>

An operation in mathematics is a function that takes zero or more input values and produces a well-defined output value.
The operation's arity is determined by the number of operands.
Similar to the principle of superposition denotes a mathematical operation (e.g., differentiation, multiplication by a constant) in which the action on the sum of two functions is the sum of the action on each function.

Two numbers with HCF 6 can be:

6a, 6b where a, b are coprime numbers a, b > 1 as numbers greater than 6.

=> 6 × 60 = 6a × 6b
=> 10 = ab
ab = 10
10 = 1 × 10 = 2 × 5
a , b > 1

As a result, a and b can be 2 or 5.

The numbers are 6 × 2 and 6 × 5.

Therefore, with an HCF of 6 and an LCM of 60, the numbers 12 and 30 are greater than 6.

Know more about mathematical operations here:

brainly.com/question/18831322

#SPJ9

You might be interested in

If f(x) =2x+3 then the point (1,5) is on the graph of f true or false

ss7ja [257]

F(x)=y=2x+3
P (1,5)
Substitute x and y:
5=2*(1)+3
5=5 true
Point P (1,5) is on the graph of f.

3 0

3 years ago

Smooth, suave, ingratiatingly pleasant

mash [69]

I say the answer is B. Bland

4 0

3 years ago

(2x+10) find value for x

Finger [1]

2x+10=0
subtract 10 from both sides
2x=(-10)
divide 2 from both sides
x=-5

4 0

3 years ago

Read 2 more answers

Write the equation of a possible rational

kondor19780726 [428]

Answer:

The graph has a removable discontinuity at x=-2.5 and asymptoe at x=2, and passes through (6,-3)

Step-by-step explanation:

A rational equation is a equation where

$\frac{p(x)}{q(x)}$

where both are polynomials and q(x) can't equal zero.

1. Discovering asymptotes. We need a asymptote at x=2 so we need a binomial factor of

$(x - 2)$

in our denomiator.

So right now we have

$\frac{p(x)}{(x - 2)}$

2. Removable discontinues. This occurs when we have have the same binomial factor in both the numerator and denomiator.

We can model -2.5 as

$(2x + 5)$

So we have as of right now.

$\frac{(2x + 5)}{(x - 2)(2x + 5)}$

Now let see if this passes throught point (6,-3).

$\frac{(2x + 5)}{(x - 2)(2x + 5)} = y$

$\frac{(17)}{68} = \frac{1}{4}$

So this doesn't pass through -3 so we need another term in the numerator that will make 6,-3 apart of this graph.

If we have a variable r, in the numerator that will make this applicable, we would get

$\frac{(2x + 5)r}{(2x + 5)(x - 2)} = - 3$

Plug in 6 for the x values.

$\frac{17r}{4(17)} = - 3$

$\frac{r}{4} = - 3$

$r = - 12$

So our rational equation will be

$\frac{ - 12(2x + 5)}{(2x + 5)(x - 2)}$

$\frac{ - 24x - 60}{2 {x}^{2} + x - 10}$

We can prove this by graphing

5 0

2 years ago

The sum of 2 numbers (x and y) is 20, and their product is 51. What are the values of x and y

Murljashka [212]

Answer: 13&7

Step-by-step explanation:

The sum of two numbers x&y is 20

Product of the 2numbers xy is 51

X+y=20......equation 1

XY=51.........equation 2

X=20-y.......equation 3

Substitute for x inequation 3 into equation 2

So we have;

(20-y)y=51

Open the bracket

20y-y^2=51

Collect like terms,so we have

Y2-20y+51=0

Factorise

Factors of 51 is 13&7

So -13-7=-20 &-13×-7=51

Substitute

Y2-13y-7y+51=0

Y(y-13)-7(y-13)=0

(Y-7)(y-13)=0

Y-7=0

Y=7 or

Y-13=0

Y=13

Y=13 or 7

Substitute for y in equation 1

X+y=20

X+7=20

X=20-7

X=13 or

X+13=20

X=20-13

X=7

Therefore, X &Y = 13 &7

4 0

2 years ago