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The answer depends on whether ab represents a 2-digit number or, as in algebra, represents the product of a and b.
If ab represents a 2 digit number, so that, in fact, it is equivalent to 10*a+b, then it is divisible by 2 if b = 0, 2, 4, 6 or 8 and divisible by 5 if b is 0 or 5. The value of a makes no difference.
If ab represents the product of a and b and they are both integers, then ab is a multiple of 2 or 5 if at least one of a or b is a multiple or 2 or 5.
It gets a lot more complicated in the latter case if they can be non-integers.
The answer depends on whether ab represents a 2-digit number or, as in algebra, represents the product of a and b.
If ab represents a 2 digit number, so that, in fact, it is equivalent to 10*a+b, then it is divisible by 2 if b = 0, 2, 4, 6 or 8 and divisible by 5 if b is 0 or 5. The value of a makes no difference.
If ab represents the product of a and b and they are both integers, then ab is a multiple of 2 or 5 if at least one of a or b is a multiple or 2 or 5.
It gets a lot more complicated in the latter case if they can be non-integers.
The answer depends on whether ab represents a 2-digit number or, as in algebra, represents the product of a and b.
If ab represents a 2 digit number, so that, in fact, it is equivalent to 10*a+b, then it is divisible by 2 if b = 0, 2, 4, 6 or 8 and divisible by 5 if b is 0 or 5. The value of a makes no difference.
If ab represents the product of a and b and they are both integers, then ab is a multiple of 2 or 5 if at least one of a or b is a multiple or 2 or 5.
It gets a lot more complicated in the latter case if they can be non-integers.
The answer depends on whether ab represents a 2-digit number or, as in algebra, represents the product of a and b.
If ab represents a 2 digit number, so that, in fact, it is equivalent to 10*a+b, then it is divisible by 2 if b = 0, 2, 4, 6 or 8 and divisible by 5 if b is 0 or 5. The value of a makes no difference.
If ab represents the product of a and b and they are both integers, then ab is a multiple of 2 or 5 if at least one of a or b is a multiple or 2 or 5.
It gets a lot more complicated in the latter case if they can be non-integers.
The answer depends on whether ab represents a 2-digit number or, as in algebra, represents the product of a and b.
If ab represents a 2 digit number, so that, in fact, it is equivalent to 10*a+b, then it is divisible by 2 if b = 0, 2, 4, 6 or 8 and divisible by 5 if b is 0 or 5. The value of a makes no difference.
If ab represents the product of a and b and they are both integers, then ab is a multiple of 2 or 5 if at least one of a or b is a multiple or 2 or 5.
It gets a lot more complicated in the latter case if they can be non-integers.
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Why is 25 a multiple of 5?
25 is not a multiple of 5 - 5 is a multiple of 25. 25/5 is 5. So we know that the multiplication of 5 by 5 is 25. We also know 5 is a multiple of 25 because the square root of 25 is 5. Finally we know that 5 is a multiple of 25 because 5² is 25. Which is the inverse of the square root of 25.
What is the LCM of 80 and 50?
LCM of (a) and (b) is given as (ab)/(GCF(a,b)) To find the GCF, factor and find common factors. 50 factors to [2 5 5] 80 factors to [2 2 2 2 5] Common factors are [2 5] = 10 50 * 80 / 10 = 400 ■
What two number that multiple together equal 1000 that do not end in 0?
5 and 25*5*5*2*2*2(I know it's a long shot but there are only two differentnumbers)
How do you draw a triangle with an area of 5 square units?
== == 1) Draw a line segment AB of 5 units 2) Draw the perpendicular bisector CD of AB such that Cd meerts AB at C. 3) Mark off CE = 2 units on CD 4) Draw the straight line segments AE & BE. ABE is your triangle. Its base (AB) = 5 and height (CE) = 2, so its area = [base x ht] / 2 = 5 sq units
What is the midpoint of AB if A is 9 and 8 and B is 5 and 2?
It is: (9+5)/2 and (8+2)/2 which is 7 and 5 Midpoint: (7, 5)
What is the measure of line segment AB if A is -5 and B is -3?
-2
What is the least common multiple for 3 and 5 and 2?
The least common multiple of 3 and 5 is 15. This, however is not a multiple of 2 so we must find the next common multiple of 3 and 5. This is 30. Since 30 is also a multiple of 2, 30 is therefore the least common multiple of 3, 5 and 2.
Why is 25 a multiple of 5?
25 is not a multiple of 5 - 5 is a multiple of 25. 25/5 is 5. So we know that the multiplication of 5 by 5 is 25. We also know 5 is a multiple of 25 because the square root of 25 is 5. Finally we know that 5 is a multiple of 25 because 5² is 25. Which is the inverse of the square root of 25.
What is the LCM of 80 and 50?
LCM of (a) and (b) is given as (ab)/(GCF(a,b)) To find the GCF, factor and find common factors. 50 factors to [2 5 5] 80 factors to [2 2 2 2 5] Common factors are [2 5] = 10 50 * 80 / 10 = 400 ■
What two number that multiple together equal 1000 that do not end in 0?
5 and 25*5*5*2*2*2(I know it's a long shot but there are only two differentnumbers)
What is AB equals C plus D?
AB = 2 x 4 (8) C = 3 D = 5
If A (-2 -4) and B (-8 4) what is the length of Ab?
Using the distance formula the length of ab is 5 units
How do you draw a triangle with an area of 5 square units?
== == 1) Draw a line segment AB of 5 units 2) Draw the perpendicular bisector CD of AB such that Cd meerts AB at C. 3) Mark off CE = 2 units on CD 4) Draw the straight line segments AE & BE. ABE is your triangle. Its base (AB) = 5 and height (CE) = 2, so its area = [base x ht] / 2 = 5 sq units
Is 4725 a multiple of 5?
yes because if it has a zero or five at the end you know its a multiple of 5
What is the least common multiple 2 and 5?
Least Common Multiple (LCM) for 2 5 is 10
What is the midpoint of AB if A is 9 and 8 and B is 5 and 2?
It is: (9+5)/2 and (8+2)/2 which is 7 and 5 Midpoint: (7, 5)
How many ways can you arrange 5 items if two have to be next to each other?
There are 48 possibilities where 2 of 5 items must be adjacent.If you have 5 items, and they can be arranged in any order, there are 1205 x 4 x 3 x 2 ways to arrange them, for example ABCDE, ABCED, and so forth.However, if any two need to be placed next to each other, the number of variations is reduced to [4 x 3 x 2] x2 (=48), where there are only 4 separate "units" arranged, but the double-unit can appear with either of the pair first.For example, if A and B must be together, you have 24 possibilities:(AB)CDE(AB)CED(AB)DCE(AB)DEC(AB)EDC(AB)ECDC(AB)DEC(AB)EDD(AB)CED(AB)ECE(AB)CDE(AB)DCCD(AB)EDC(AB)ECE(AB)DEC(AB)DDE(AB)CED(AB)CCDE(AB)CED(AB)DEC(AB)DCE(AB)EDC(AB)ECD(AB)and another 24 where (AB) is replaced by (BA).
