All of these involve simple substitution. Just replace the variable with the actual value.
8z + 3 >>> 8(8) + 3 = 64 + 3 = 67
3(7+ x^2) >>> 3(7 + 2^2) = 3(7 + 4) = 3(11)=33
s - 5t + s^2 >>> 4 - 5(-1) + 4^2 = 4 + 5 + 16
Get the trend? I am not sure I get that last one, is it:
x - y*3^2?
Answer:
1.23 x 10^8
hope this helps
have a good day :)
Step-by-step explanation:
11x+3y=103 and y=3x+1.
11x+3(3x+1)=103 -----> plug in (3x+1) for y in the first equation. You will want to distribute the 3 to the 3x+1 to get something that looks like:
11x+9x+3=103 ------> now you want to combine like terms
20x+3=103 ---> subtract 3 from both sides
20x=100 ----> divide both sides by 20
x=5
y=3(5)+1 ---> I like to plug in this to the equation that already has y isolated. 3*5 is 15, add 1 and you find that y=16.
(5, 16) will be your final answer (:
Step-by-step explanation:
Starting with the first one, we know that 3²=9 and 4²=16, so √10 is between 3 and 4. Similarly, √27 is a little over 5. √3 is close to 2, √2 is close to 1, and √9=3, so for simplicity, we can write this as 
The 3.something and the 3 kind of cross out, and we're left with 5.something over 2 1.somethings, which can be closer to 2 or 1, which is somewhat unclear -- therefore, our answer can be D or E, and we'll wait on this one
For the second one, we can cross out the √6s to get 
-- π is a little greater than 3, so this is E, making the first one D
For the third one, we can cross out the √3 to get (3-2)/2 = 1/2, or A as it is between 0 and 1
For the last one, note the multiplication -- cross out the √3 to get (3*2√2)/6, and we get 6√2/6=√2, or C
