Discover the answers you need at Westonci.ca, where experts provide clear and concise information on various topics. Explore a wealth of knowledge from professionals across different disciplines on our comprehensive platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
Sure, let's solve the expression \(\frac{4r - t + s}{2}\) step-by-step given that \( r = 4 \), \( t = 1 \), and \( s = 2 \).
1. Substitute the given values into the expression:
[tex]\[ \frac{4r - t + s}{2} \rightarrow \frac{4 \cdot 4 - 1 + 2}{2} \][/tex]
2. Calculate the value inside the numerator first:
\( 4 \cdot 4 = 16 \)
So the expression now is:
[tex]\[ \frac{16 - 1 + 2}{2} \][/tex]
3. Perform the subtraction and addition in the numerator:
\( 16 - 1 = 15 \)
\( 15 + 2 = 17 \)
Now the expression is:
[tex]\[ \frac{17}{2} \][/tex]
4. Finally, perform the division:
[tex]\[ \frac{17}{2} = 8.5 \][/tex]
So, the value of [tex]\(\frac{4r - t + s}{2}\)[/tex] when [tex]\( r = 4 \)[/tex], [tex]\( t = 1 \)[/tex], and [tex]\( s = 2 \)[/tex] is [tex]\( 8.5 \)[/tex].
1. Substitute the given values into the expression:
[tex]\[ \frac{4r - t + s}{2} \rightarrow \frac{4 \cdot 4 - 1 + 2}{2} \][/tex]
2. Calculate the value inside the numerator first:
\( 4 \cdot 4 = 16 \)
So the expression now is:
[tex]\[ \frac{16 - 1 + 2}{2} \][/tex]
3. Perform the subtraction and addition in the numerator:
\( 16 - 1 = 15 \)
\( 15 + 2 = 17 \)
Now the expression is:
[tex]\[ \frac{17}{2} \][/tex]
4. Finally, perform the division:
[tex]\[ \frac{17}{2} = 8.5 \][/tex]
So, the value of [tex]\(\frac{4r - t + s}{2}\)[/tex] when [tex]\( r = 4 \)[/tex], [tex]\( t = 1 \)[/tex], and [tex]\( s = 2 \)[/tex] is [tex]\( 8.5 \)[/tex].
Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.