Answer :
Let's solve the given problem step by step.
1. Calculate [tex]\(a + b\)[/tex]:
Given values:
[tex]\[a = 2\][/tex]
[tex]\[b = 4\][/tex]
First, we need to find the sum of [tex]\(a\)[/tex] and [tex]\(b\)[/tex]:
[tex]\[a + b = 2 + 4 = 6\][/tex]
So, [tex]\((a + b) = 6\)[/tex].
2. Calculate [tex]\(c \cdot d\)[/tex]:
Given values:
[tex]\[c = 3\][/tex]
[tex]\[d = 12\][/tex]
Now, let's multiply [tex]\(c\)[/tex] and [tex]\(d\)[/tex]:
[tex]\[c \cdot d = 3 \cdot 12 = 36\][/tex]
3. Find the square root of [tex]\(c \cdot d\)[/tex]:
We need to find the square root of [tex]\(c \cdot d = 36\)[/tex]:
[tex]\[\sqrt{36} = 6.0\][/tex]
So, [tex]\(\sqrt{c \cdot d} = 6.0\)[/tex].
4. Multiply [tex]\((a + b)\)[/tex] by [tex]\(\sqrt{c \cdot d}\)[/tex]:
We have already calculated [tex]\((a + b) = 6\)[/tex] and [tex]\(\sqrt{c \cdot d} = 6.0\)[/tex]. Now, we need to multiply these two results together:
[tex]\[(a + b) \sqrt{c \cdot d} = 6 \times 6.0 = 36.0\][/tex]
Therefore, the value of [tex]\((a + b) \sqrt{c \cdot d}\)[/tex] is [tex]\(\boxed{36.0}\)[/tex].
1. Calculate [tex]\(a + b\)[/tex]:
Given values:
[tex]\[a = 2\][/tex]
[tex]\[b = 4\][/tex]
First, we need to find the sum of [tex]\(a\)[/tex] and [tex]\(b\)[/tex]:
[tex]\[a + b = 2 + 4 = 6\][/tex]
So, [tex]\((a + b) = 6\)[/tex].
2. Calculate [tex]\(c \cdot d\)[/tex]:
Given values:
[tex]\[c = 3\][/tex]
[tex]\[d = 12\][/tex]
Now, let's multiply [tex]\(c\)[/tex] and [tex]\(d\)[/tex]:
[tex]\[c \cdot d = 3 \cdot 12 = 36\][/tex]
3. Find the square root of [tex]\(c \cdot d\)[/tex]:
We need to find the square root of [tex]\(c \cdot d = 36\)[/tex]:
[tex]\[\sqrt{36} = 6.0\][/tex]
So, [tex]\(\sqrt{c \cdot d} = 6.0\)[/tex].
4. Multiply [tex]\((a + b)\)[/tex] by [tex]\(\sqrt{c \cdot d}\)[/tex]:
We have already calculated [tex]\((a + b) = 6\)[/tex] and [tex]\(\sqrt{c \cdot d} = 6.0\)[/tex]. Now, we need to multiply these two results together:
[tex]\[(a + b) \sqrt{c \cdot d} = 6 \times 6.0 = 36.0\][/tex]
Therefore, the value of [tex]\((a + b) \sqrt{c \cdot d}\)[/tex] is [tex]\(\boxed{36.0}\)[/tex].