All categories

4 years ago

Can two numbers have a GCF that is bigger than their LCM? Explain how you know?

1 answer:

3 0

The answer to this question would be: No, they can't

Greatest common factor biggest possible value is the number value itself. There is no factor that bigger than the number.

But the least common multiple smallest possible value is the number. There is no multiple that smaller than the number.

So it is not possible to have GCF bigger than LCM.

You might be interested in

Write a fraction that is less than 5/6 and has a denominator of 8

KatRina [158]

You'll want to make a common denominator with 6 and 8.
That denominator would be 24.
24/6=4 so you would have to multiply 5/6 by 4/4 to get 20/24.
Next, 24/8=3 so 1/8 could be multiplied by 3/3 to get 3/24.
Since 3 is less than 20, 1/8 is smaller than 5/6.

If you want the same numerator, 5/8 = 15/24. This would make 5/8 smaller than 5/6 as well.
The highest eighth you can go is 6/8 which is 18/24.
So you can use any numerator between 1 and 6 with a denominator of 8 to get a fraction smaller than 5/6.

3 0

3 years ago

A regular pentagon ABCDE is shown work out angle x

kow [346]

Answer:

108 degrees inside each corner

Step-by-step explanation:

360/5=72

180-72=108

5 0

3 years ago

Which equation is y = –6x^2 + 3x + 2 rewritten in vertex form? y = negative 6 (x minus 1) squared + 8 y = negative 6 (x + one-fo

mart [117]

Answer:

$y = -6(x - \frac{1}{2})^2 -\frac{7}{2}$

Step-by-step explanation:

Given:

$y = -6x^2 + 3x + 2$

Required

Rewrite in vertex form

The vertex form of an equation is in form of: $y = a(x - h)^2+ k$

Solving: $y = -6x^2 + 3x + 2$

Subtract 2 from both sides

$y - 2 = -6x^2 + 3x + 2 - 2$

$y - 2 = -6x^2 + 3x$

Factorize expression on the right hand side by dividing through by the coefficient of x²

$y - 2 = -6(x^2 + \frac{3x}{-6})$

$y - 2 = -6(x^2 - \frac{3x}{6})$

$y - 2 = -6(x^2 - \frac{x}{2})$

Get a perfect square of coefficient of x; then add to both sides

------------------------------------------------------------------------------------

<em>Rough work</em>

The coefficient of x is $\frac{-1}{2}$

It's square is $(\frac{-1}{2})^2 = \frac{1}{4}$

Adding inside the bracket of $-6(x^2 - \frac{x}{2})$ to give: $-6(x^2 - \frac{x}{2} + \frac{1}{4})$

To balance the equation, the same expression must be added to the other side of the equation;

Equivalent expression is: $-6(\frac{1}{4})$

------------------------------------------------------------------------------------

The expression becomes

$y - 2 -6(\frac{1}{4})= -6(x^2 - \frac{x}{2} + \frac{1}{4})$

$y - 2 -\frac{6}{4}= -6(x^2 - \frac{x}{2} + \frac{1}{4})$

$y - 2 -\frac{3}{2}= -6(x^2 - \frac{x}{2} + \frac{1}{4})$

Factorize the expression on the right hand side

$y - 2 -\frac{3}{2}= -6(x - \frac{1}{2})^2$

$y - (2 +\frac{3}{2})= -6(x - \frac{1}{2})^2$

$y - (\frac{4+3}{2})= -6(x - \frac{1}{2})^2$

$y - (\frac{7}{2})= -6(x - \frac{1}{2})^2$

$y +\frac{7}{2} = -6(x - \frac{1}{2})^2$

Make y the subject of formula

$y = -6(x - \frac{1}{2})^2 -\frac{7}{2}$

<em>Solved</em>

7 0

3 years ago

Read 2 more answers

Find the locus of the point (x,y) which is a distance 3 from the point (-2, 1)

Reptile [31]

Consider this option:
1. rule: Circle is the locus of the points which are on the same distance from the given point. Common view of the equation is: (x-a)²+(y-b)²=r², where r - the distance from the given point to the described locus, a&b - coordinates of the given point.
2. using item 1:
(x+2)²+(y-1)²=3² or (x+2)²+(y-1)²=9.

6 0

3 years ago

Question 8b , thanks

marta [7]

Answer:

Step-by-step explanation:

$a = \frac{2b+1}{3b-1}$