Okay, so on number one I have no idea. I keep getting 20 because 14 - (-3)2 = 14- (-6) which is really 14 + 6=20.
Number 2 is r+15=61
Number 3 I didn't get any of those answers because you have to solve what is in the parenthesis first so3*5= 15+1=16 then you multiply that by two giving you 16*2=32 and 42-32=10
<h3>Given</h3>
P = 3r + 2s
<h3>Find</h3>
the corresponding equation for s
<h3>Solution</h3>
First of all, look at how this is evaluated in terms of what happens to a value for s.
- s is multiplied by 2
- 3r is added to that product
To solve for s, you undo these operations in reverse order. The "undo" for addition is adding the opposite. The "undo" for multiplication is division (or multiplication by the reciprocal).
... P = 3r + 2s . . . . . . starting equation
... P - 3r = 2s . . . . . . -3r is added to both sides to undo addition of 3r
... (P -3r)/2 = s . . . . . both sides are divided by 2 to undo the multiplication
Note that the division is of everything on both sides of the equation. That is why we need to add parentheses around the expression that was on the left—so the whole thing gets divided by 2.
Your solution is ...
... s = (P - 3r)/2
The solution to the system of equation are x=2, y=0, z=6
<h3>System of equations</h3>
System of equations are equations that contains unknown variables.
Given the equations
3x+y+2z=8
8y+6z=36
12y+2z=12
From equation 2 and 3
8y+6z=36 * 1
12y+2z=12 * 3
______________
8y+6z=36
36y+6z= 36
Subtract
8y - 36y = 36 - 36
-28y =0
y = 0
Substitute y = 0 into equation 2
8(0)+6z=36
6z = 36
z = 6
From equation 1
3x+y+2z =8
3x + 0 + 2(6) = 8
3x = 8 - 12
3x = 6
x = 2
Hence the solution to the system of equation are x=2, y=0, z=6
Learn more on system of equation here: brainly.com/question/14323743
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Answer:
10
Step-by-step explanation:
20-(5×2)
5×2=10
20-10=10