Answer:
-6
Step-by-step explanation:
We use PEMDAS to solve this,
so P stands for parentheses, so that's where we start.
We first, square the innermost parentheses with the exponent which is the E in PEMDAS, then then the outer parentheses
-12/3*(-8+16-6)+2
-12/3*(2)+2
Now we divide as in Division in PEMDAS.
-4*2+2
Now we multiply as in Multiplication in PEMDAS.
-8+2
Now we add as in A for Addition
-6
In PEMDAS, Multiplication doesn't always come before division, and same for addition and subtraction.
Time-and-a-half doesn't make enough sense to answer the question. If you fix it, or comment below the correction I will gladly help.
Subtraction
multiplication
yayyyyyyyyyyy
Idek hun I don’t see anything
Answer:
option 4.
16 square units
Step-by-step explanation:
as we do not have the measures of the sides, but if the points of the vertices with Pythagoras we can calculate the sides.
P = (2 , 4)
S = (4 , 2)
we have to subtract the values of p from s
PS = (4 - 2 , 2 - 4)
PS = (2 , -2)
by pitagoras h ^ 2 = c1 ^ 2 + c2 ^ 2
h: hypotenuse
c1: leg 1
c2: leg 2
PS^2 = 2^2 + -2^2
PS = √ 4 + 4
PS = √8
PS = 2√2
S = (4 , 2)
R = (8 , 6)
SR = (8-4 , 6-2)
SR = (4 , 4)
by pitagoras h ^ 2 = c1 ^ 2 + c2 ^ 2
h: hypotenuse
c1: leg 1
c2: leg 2
SR^2 = 4^2 + 4^2
SR = √ (16 + 16)
SR = √32
SR = 4√2
having the values of 2 of its sides we multiply them and obtain their area
PS * RS = Area
2√2 * 4√2 =
16
