Find the HCF and LCM by prime factorization method 112 and 150

lara31 [8.8K]

Well, first of all, the first statement (ABC = ADC) looks like it just says
that the two halves of the little square ... each side of the diagonal ...
are congruent. That's no big deal, and it's no help in answering the
question.

The effect of the dilation is that all the DIMENSIONS of the square
are doubled ... each side of the square becomes twice as long.

Then, when you multiply (length x width) to get the area, you'd have

Area = (2 x original length) x (2 x original width)

and that's
the same as (2 x 2) x (original length x original width)

= (4) x (original area) .

Here's an easy, useful factoid to memorize:

-- Dilate a line (1 dimension) by 'x' times . . . multiply the length by x¹

-- Dilate a shape (2 dimensions) by 'x' . . . multiply area by x²

-- Dilate a solid (3 dimensions) by 'x' . . . multiply volume by x³

And that's all the dimensions we have in our world.
_______________________________

Oh, BTW . . .

-- Dilate a point (0 dimensions) by 'x' . . . multiply it by x⁰ (1)

5 0

3 years ago

What do you think is true of a line and a point not on the line? Explain

statuscvo [17]

The point is not a solution to the equation of the line.

Hope this helps!

3 0

3 years ago

Y varies directly as x. Y=42 when x=6. Find the value of y when x =15.

pentagon [3]

Answer:

y=120

Step-by-step explanation:

Y varies directly as x is written as

$y \alpha x$

introducing a constant

y=kx

from the question, when y=48, x=6

substitute it in the formula

48=6k

making k the subject by dividing through by 6

$\frac{48}{6} = \frac{6k}{6}$

k=48/6

k=8

put the value of k in the expression y=kx

y=8x

from the question,the value of y when x=15 is given by

y=8×15

y=120.

4 0

3 years ago

Can any one help me solve liner eqations?

sveticcg [70]

S I can
which linear equation due u want?

3 0

3 years ago

How do I solve this?

Crazy boy [7]

Answer:

71°

142°

Step-by-step explanation:

$\because QS \: is \: the \: bisector \: of \: \angle \: PQR\\\therefore m\angle RQS =m\angle PQS\\\because m\angle RQS = 71\degree...(given) \\ \huge \red{ \boxed{\therefore m\angle PQS = 71\degree}} \\ \because \: m\angle \: PQR = m\angle PQS + m\angle RQS \\ \because \: m\angle \: PQR = 71\degree + 71\degree \\ \huge \purple{ \boxed{\therefore \: m\angle \: PQR = 142\degree}}$

7 0

3 years ago

Find the HCF and LCM by prime factorization method 112 and 150​

Find the HCF and LCM by prime factorization method 112 and 150