Complete the simplification below by finding the values of the capitalised pronumerals.๐ is a negative number.ย 6(2๐โ4๐)โ6(6๐โ4๐)+3(4๐โ8๐)4=๐(2๐+๐๐+๐๐)ย ๐=๐=
Question
Complete the simplification below by finding the values of the capitalised pronumerals.๐ is a negative number.ย 6(2๐โ4๐)โ6(6๐โ4๐)+3(4๐โ8๐)4=๐(2๐+๐๐+๐๐)ย ๐=๐=
Solution
The question seems to be incomplete. You have asked for the values of X, Y, and Z but you have not provided any specific values for a, b, and c.
However, we can simplify the equation and express Y and Z in terms of X.
First, let's simplify the left side of the equation:
6(2a-4c) - 6(6a-4b) + 3(4b-8c) = 12a - 24c - 36a + 24b + 12b - 24c This simplifies to: -24a + 36b - 48c
Now, let's equate this to the right side of the equation:
-24a + 36b - 48c = X(2a + Yb + Zc)
Now, we can compare the coefficients on both sides of the equation to find the values of X, Y, and Z.
Comparing the coefficients of 'a', we get X*2 = -24, so X = -24/2 = -12.
Comparing the coefficients of 'b', we get X*Y = 36, so Y = 36/X = 36/-12 = -3.
Comparing the coefficients of 'c', we get X*Z = -48, so Z = -48/X = -48/-12 = 4.
So, X = -12, Y = -3, and Z = 4.
Similar Questions
Simplify the following expression. -6โข(12โข๐ฅ-16โข๐ฆ)+6โข(-12โข๐ฅ-56โข๐ฆ)=
Assignment:1.2.3.(๐ท2 + 5๐ท + 6)๐ฆ = 0โฎ ๐ฆ = ๐ถ1๐โ2๐ฅ + ๐ถ2๐โ3๐ฅ(2๐ท2 โ 5๐ท โ 3)๐ฆ = 0โฎ ๐ฆ = ๐ถ1๐3๐ฅ + ๐ถ2๐โ ๐ฅ2(๐ท4 โ 6๐ท + 1)๐ฆ = 0โฎ ๐ฆ = ๐ถ1 + ๐ถ2๐๐ฅ + (๐ถ3 + ๐ถ4๐ฅ)๐๐ฅโฎ ๐ฆ = ๐ถ1๐โ2๐ฅ + ๐ถ2๐โ3๐ฅโฎ ๐ฆ = ๐ถ1๐3๐ฅ + ๐ถ2๐โ ๐ฅ2โฎ ๐ฆ = ๐ถ1 + ๐ถ2๐๐ฅ + (๐ถ3 + ๐ถ4๐ฅ)๐๐ฅ
Multiply: 6โข(๐ฅ-1)= โ1
Expand and simplify the following expression:6(๐ฅ+7)โ3(4๐ฅโ2)
๐(๐ฅ) = ๐ฅ3 + 3๐ฅ โ 4
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