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Calculate the values of [tex]\( c \)[/tex] and [tex]\( d \)[/tex] in the expression [tex]\( 7 \sqrt{12} \times \sqrt{5} \)[/tex] given in its simplest form as [tex]\( c \sqrt{d} \)[/tex], where [tex]\( c \)[/tex] and [tex]\( d \)[/tex] are integers.

Sagot :

GinnyAnswer
Sure, let's break down the steps to simplify the given expression [tex]\(7 \sqrt{12} \times \sqrt{5}\)[/tex] into the form [tex]\(c \sqrt{d}\)[/tex] where [tex]\(c\)[/tex] and [tex]\(d\)[/tex] are integers.

1. Combine the square roots:

The given expression is [tex]\(7 \sqrt{12} \times \sqrt{5}\)[/tex]. We know that multiplying square roots can be combined under one square root:
[tex]\[ \sqrt{12} \times \sqrt{5} = \sqrt{12 \times 5} = \sqrt{60} \][/tex]

2. Simplify the square root:

Next, we simplify [tex]\(\sqrt{60}\)[/tex]. We can factorize 60 to its prime factors:
[tex]\[ 60 = 2 \times 2 \times 3 \times 5 = 2^2 \times 3 \times 5 \][/tex]
Using the property of square roots, we can separate the perfect square (which is [tex]\(2^2\)[/tex]):
[tex]\[ \sqrt{60} = \sqrt{2^2 \times 3 \times 5} = \sqrt{2^2} \times \sqrt{3 \times 5} = 2 \times \sqrt{15} \][/tex]
Thus, [tex]\(\sqrt{60}\)[/tex] simplifies to [tex]\(2 \sqrt{15}\)[/tex].

3. Multiply by the constant:

We now multiply the result by 7:
[tex]\[ 7 \sqrt{12} \times \sqrt{5} = 7 \times 2 \sqrt{15} = 14 \sqrt{15} \][/tex]

4. Identify [tex]\(c\)[/tex] and [tex]\(d\)[/tex]:

Comparing our result with the form [tex]\(c \sqrt{d}\)[/tex], we see that [tex]\(c = 14\)[/tex] and [tex]\(d = 15\)[/tex].

Therefore, the values of [tex]\(c\)[/tex] and [tex]\(d\)[/tex] are [tex]\(c = 14\)[/tex] and [tex]\(d = 15\)[/tex].
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