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2 years ago

If ‘a ’ is a factor of ‘b ’, then the H.C.F. of a and b is

Mathematics

1 answer:

Nookie1986 [14]2 years ago

5 0

Answer:

Prime Factors, HCF and LCM

We say that d is a common factor of a and b if d is a factor of both a and b. We say that d is the highest common factor of a and b if there is no other factor of both a and b that is greater than d. Notation: HCF(a,b) = “highest common factor of a and b” .

mark me brainliest..

hope it helps uh!

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Need help answering question 4

raketka [301]

Answer:

(-1,5)

Step-by-step explanation:

To find this, you have to add both "x" coordinates, then divide them by two.

Same thing goes for the "y" coordinates, add them, then divide them by two.

x-coordinates ---> 2+(-4)= -2

y-coordinates ---> 1+9= 10

-2÷2= -1

10÷2= 5

That gives you, (-1,5).

4 0

2 years ago

In the diagram below, KL = 12 and LM = 8. Which additional facts would guarantee that JKLM is a parallelogram? Check all that ap

Liono4ka [1.6K]

Answer:

A. JK = 12, JK is parallel to LM

D. JK = 8, JM = 12

Step-by-step explanation:

Based on the given figure above, the additional facts that would guarantee that JKLM is a parallelogram are answers A and D.

By definition, a parallelogram is a figure that has opposite sides parallel or it has two pairs of sides that are parallel.

8 0

2 years ago

Carter draws one side of equilateral △PQR on the coordinate plane at points P(-3,2) and Q(5,2). Which ordered pair is a possible

Law Incorporation [45]

Step-by-step explanation:

Hey, there!!!

Let me simply explain you about it.

We generally use the distance formula to get the points.

let the point R be (x,y)

As it an equilateral triangle it must have equal distance.

now,

let's find the distance of PQ,

we have, distance formulae is;

$pq = \sqrt{( {x2 - x1)}^{2} + ( {y2 - y1)}^{2} }$

$or \: \sqrt{( {5 + 3)}^{2} + ( {2 - 2)}^{2} }$

By simplifying it we get,

$8$

Now,

again finding the distance between PR,

$pr = \sqrt{( {x2 - x1}^{2} + ( {y2 - y1)}^{2} }$

or,

$\sqrt{( {x + 3)}^{2} + ( {y - 2)}^{2} }$

By simplifying it we get,

$= \sqrt{ {x}^{2} + {y}^{2} + 6x - 4y + 13 }$

now, finding the distance of QR,

$qr = \sqrt{( {x - 5)}^{2} + ( {y - 2)}^{2} }$

or, by simplification we get,

$\sqrt{ {x}^{2} + {y}^{2} - 10x - 4y + 29 }$

now, equating PR and QR,

$\sqrt{ {x}^{2} + {y}^{2} + 6x - 4y + 13} = \sqrt{ {x}^{2} + {y}^{2} - 10x - 4y + 29 }$

we cancelled the root ,

${x}^{2} + {y}^{2} + 6x - 4y + 13 = {x}^{2} + {y}^{2} -10x - 4y + 29$

or, cancelling all like terms, we get,

6x+13= -10x+29

16x=16

x=16/16

Therefore, x= 1.

now,

equating, PR and PQ,

$\sqrt{ {x}^{2} + {y}^{2} + 6x - 4y + 13 } = 8}$

cancel the roots,

${x}^{2} + {y}^{2} + 6x - 4y + 13 = 8$

now,

(1)^2+ y^2+6×1-4y+13=8

or, 1+y^2+6-4y+13=8

y^2-4y+13+6+1=8

or, y(y-4)+20=8

or, y(y-4)= -12

either, or,

y= -12 y=8

Therefore, y= (8,-12)

by rounding off both values, we get,

x= 1

y=(8,-12)

So, i think it's (1,8) is your answer..

Hope it helps...

3 0

3 years ago

Please, it’s due tonight and I suck at proofs:((

Slav-nsk [51]

Answer:

The question ?

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

A triangle has sides with lengths of 4 inches, 14 inches, and 17 inches. Is it a right triangle?

irga5000 [103]

Yes it is because 2 sides would be the legs while 1 side is the hyptonuese so in this case 4 is a leg and 14 and 17 is the hyptonuese

5 0

2 years ago

If ‘a ’ is a factor of ‘b ’, then the H.C.F. of a and b is​

If ‘a ’ is a factor of ‘b ’, then the H.C.F. of a and b is