Top Answer

The "triple product" is defined for vectors. You can't calculate a triple product if you don't know the components of the vectors (or some other information, that allows you to calculate those).

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-(6a)

A "Pythagorean Triple" is a set of positive integers, a, b and c that fits the rule: a2 + b2 = c2. Example: The smallest Pythagorean Triple is 3, 4 and 5

21a - 6a

2

(a-b) (a+b) = a2+b2

If by "triple dot product" you mean uÂ·vÂ·w, then no, because that would imply the existence of a dot product between a vector and a scalar.

Divide by 6a: 6a(a + 3b)

36a2 - 60a + 25 = 36a2 - 30a - 30a + 25 = 6a(6a - 5) - 5(6a - 5) = (6a - 5)(6a - 5) = (6a - 5)2

12a2b + 6a = 6a(2ab + 1)

-6a + 8 = 2 -6a = -6 a = 1

18

Suppose you have two sets of n-numbers: {a1, a2, a3, ... , an} and {b1, b2, b3, ... , bn} Then the form for the standard sum of product is a1*b1 + a2+b2 + a3*b3 + ... + an*bn

6a

6a

a+6a+a=8a

15n

What is the tonnage of york a AC model CM30-6a

6a2

2 x 3 x a = 6a

6a + 6b = 6(a + b)

36

10a-37=6a+51 -6a -6a subtract 6a from both sides and the 6a's cancel 4a-37=51 +37 +37 4a=88 /4 /4 a=22

You can do it lots of ways. You can think of it as taking 10% away or just getting 90% of the original value. Assuming your value was in cell B2, any of these would work:=B2-B2*10%=B2*90%=B2-B2*0.1=B2*0.9=B2/10*9You can do it lots of ways. You can think of it as taking 10% away or just getting 90% of the original value. Assuming your value was in cell B2, any of these would work:=B2-B2*10%=B2*90%=B2-B2*0.1=B2*0.9=B2/10*9You can do it lots of ways. You can think of it as taking 10% away or just getting 90% of the original value. Assuming your value was in cell B2, any of these would work:=B2-B2*10%=B2*90%=B2-B2*0.1=B2*0.9=B2/10*9You can do it lots of ways. You can think of it as taking 10% away or just getting 90% of the original value. Assuming your value was in cell B2, any of these would work:=B2-B2*10%=B2*90%=B2-B2*0.1=B2*0.9=B2/10*9You can do it lots of ways. You can think of it as taking 10% away or just getting 90% of the original value. Assuming your value was in cell B2, any of these would work:=B2-B2*10%=B2*90%=B2-B2*0.1=B2*0.9=B2/10*9You can do it lots of ways. You can think of it as taking 10% away or just getting 90% of the original value. Assuming your value was in cell B2, any of these would work:=B2-B2*10%=B2*90%=B2-B2*0.1=B2*0.9=B2/10*9You can do it lots of ways. You can think of it as taking 10% away or just getting 90% of the original value. Assuming your value was in cell B2, any of these would work:=B2-B2*10%=B2*90%=B2-B2*0.1=B2*0.9=B2/10*9You can do it lots of ways. You can think of it as taking 10% away or just getting 90% of the original value. Assuming your value was in cell B2, any of these would work:=B2-B2*10%=B2*90%=B2-B2*0.1=B2*0.9=B2/10*9You can do it lots of ways. You can think of it as taking 10% away or just getting 90% of the original value. Assuming your value was in cell B2, any of these would work:=B2-B2*10%=B2*90%=B2-B2*0.1=B2*0.9=B2/10*9You can do it lots of ways. You can think of it as taking 10% away or just getting 90% of the original value. Assuming your value was in cell B2, any of these would work:=B2-B2*10%=B2*90%=B2-B2*0.1=B2*0.9=B2/10*9You can do it lots of ways. You can think of it as taking 10% away or just getting 90% of the original value. Assuming your value was in cell B2, any of these would work:=B2-B2*10%=B2*90%=B2-B2*0.1=B2*0.9=B2/10*9

(-8 + b2) - (5 + b2) = -8 + b2 - 5 - b2 = -13

Magnesium sulfide (MgS) is formed by the reaction of magnesium with sulfur.