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Sagot :
Certainly! We'll take each part of your query step-by-step.
### Part 41. a) Simplify:
Given expression: [tex]\(\left(\frac{2}{15} + 2 \frac{2}{5}\right) + \left(\frac{4}{20}x + \frac{2}{2}\right) + 1 = \frac{4}{4}\)[/tex]
Let's break it down step-by-step.
1. Simplify each fraction and mixed number:
[tex]\[ \frac{2}{15} + 2 \frac{2}{5} \][/tex]
Recall that [tex]\(2 \frac{2}{5} = 2 + \frac{2}{5}\)[/tex], so:
[tex]\[ 2 + \frac{2}{5} = \frac{2 \cdot 5 + 2}{5} = \frac{10 + 2}{5} = \frac{12}{5} \][/tex]
Now adding [tex]\(\frac{2}{15}\)[/tex]:
[tex]\[ \frac{2}{15} + \frac{12}{5} \][/tex]
To add these, find a common denominator, which is 15:
[tex]\[ \frac{12 \cdot 3}{5 \cdot 3} = \frac{36}{15} \][/tex]
Adding the fractions:
[tex]\[ \frac{2}{15} + \frac{36}{15} = \frac{38}{15} \][/tex]
2. Simplify [tex]\(\frac{4}{20}x + \frac{2}{2}\)[/tex]:
[tex]\[ \frac{4}{20}x = \frac{1}{5}x \][/tex]
And:
[tex]\[ \frac{2}{2} = 1 \][/tex]
So, combining these:
[tex]\[ \frac{1}{5}x + 1 \][/tex]
3. Put it all together:
[tex]\[ \left( \frac{38}{15} \right) + \left( \frac{1}{5}x + 1 \right) + 1 \][/tex]
Simplifying within each term:
[tex]\[ \frac{38}{15} + 1 + 1 = \frac{38}{15} + 2 = \frac{38}{15} + \frac{30}{15} = \frac{68}{15} \][/tex]
Therefore:
[tex]\[ \frac{68}{15} + \frac{1}{5}x = \frac{4}{4} = 1 \][/tex]
So, the simplified form: [tex]\(\left(\frac{68}{15} + \frac{1}{5}x = 1\right)\)[/tex].
### Part 41. b) Make [tex]\(a\)[/tex] the subject:
Since a specific equation is missing from this part, let's assume the generic context in which we make [tex]\(a\)[/tex] the subject of the formula.
Let's consider an example equation:
[tex]\[ b = a + c \][/tex]
To make [tex]\(a\)[/tex] the subject:
[tex]\[ a = b - c \][/tex]
If you have a specific equation you need solved, please provide it.
### Part 41. c) Factorize [tex]\(y^2 - 60 + 2 - 1\)[/tex]:
Simplify the expression:
[tex]\[ y^2 - 59 - 1 = y^2 - 58 \][/tex]
Unfortunately, [tex]\(y^2 - 58\)[/tex] does not factor nicely into integers since 58 is not a perfect square.
### Solve for the Gandoloundos [tex]\(m + \pi = 5 + m\)[/tex]:
Let's solve for [tex]\(m\)[/tex]:
Given:
[tex]\[ m + \pi = 5 + m \][/tex]
Subtract [tex]\(m\)[/tex] from both sides:
[tex]\[ \pi = 5 \][/tex]
This essentially means that, considering the value of [tex]\(m\)[/tex] on both sides cancels out, the equation represents a specific relationship for [tex]\(\pi\)[/tex] which is given numerically as [tex]\(\pi - 5\)[/tex].
So the solution maintains the relation showcasing constant properties, but simplifies expressing [tex]\(\pi = 5\)[/tex].
### Part 41. a) Simplify:
Given expression: [tex]\(\left(\frac{2}{15} + 2 \frac{2}{5}\right) + \left(\frac{4}{20}x + \frac{2}{2}\right) + 1 = \frac{4}{4}\)[/tex]
Let's break it down step-by-step.
1. Simplify each fraction and mixed number:
[tex]\[ \frac{2}{15} + 2 \frac{2}{5} \][/tex]
Recall that [tex]\(2 \frac{2}{5} = 2 + \frac{2}{5}\)[/tex], so:
[tex]\[ 2 + \frac{2}{5} = \frac{2 \cdot 5 + 2}{5} = \frac{10 + 2}{5} = \frac{12}{5} \][/tex]
Now adding [tex]\(\frac{2}{15}\)[/tex]:
[tex]\[ \frac{2}{15} + \frac{12}{5} \][/tex]
To add these, find a common denominator, which is 15:
[tex]\[ \frac{12 \cdot 3}{5 \cdot 3} = \frac{36}{15} \][/tex]
Adding the fractions:
[tex]\[ \frac{2}{15} + \frac{36}{15} = \frac{38}{15} \][/tex]
2. Simplify [tex]\(\frac{4}{20}x + \frac{2}{2}\)[/tex]:
[tex]\[ \frac{4}{20}x = \frac{1}{5}x \][/tex]
And:
[tex]\[ \frac{2}{2} = 1 \][/tex]
So, combining these:
[tex]\[ \frac{1}{5}x + 1 \][/tex]
3. Put it all together:
[tex]\[ \left( \frac{38}{15} \right) + \left( \frac{1}{5}x + 1 \right) + 1 \][/tex]
Simplifying within each term:
[tex]\[ \frac{38}{15} + 1 + 1 = \frac{38}{15} + 2 = \frac{38}{15} + \frac{30}{15} = \frac{68}{15} \][/tex]
Therefore:
[tex]\[ \frac{68}{15} + \frac{1}{5}x = \frac{4}{4} = 1 \][/tex]
So, the simplified form: [tex]\(\left(\frac{68}{15} + \frac{1}{5}x = 1\right)\)[/tex].
### Part 41. b) Make [tex]\(a\)[/tex] the subject:
Since a specific equation is missing from this part, let's assume the generic context in which we make [tex]\(a\)[/tex] the subject of the formula.
Let's consider an example equation:
[tex]\[ b = a + c \][/tex]
To make [tex]\(a\)[/tex] the subject:
[tex]\[ a = b - c \][/tex]
If you have a specific equation you need solved, please provide it.
### Part 41. c) Factorize [tex]\(y^2 - 60 + 2 - 1\)[/tex]:
Simplify the expression:
[tex]\[ y^2 - 59 - 1 = y^2 - 58 \][/tex]
Unfortunately, [tex]\(y^2 - 58\)[/tex] does not factor nicely into integers since 58 is not a perfect square.
### Solve for the Gandoloundos [tex]\(m + \pi = 5 + m\)[/tex]:
Let's solve for [tex]\(m\)[/tex]:
Given:
[tex]\[ m + \pi = 5 + m \][/tex]
Subtract [tex]\(m\)[/tex] from both sides:
[tex]\[ \pi = 5 \][/tex]
This essentially means that, considering the value of [tex]\(m\)[/tex] on both sides cancels out, the equation represents a specific relationship for [tex]\(\pi\)[/tex] which is given numerically as [tex]\(\pi - 5\)[/tex].
So the solution maintains the relation showcasing constant properties, but simplifies expressing [tex]\(\pi = 5\)[/tex].
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