Answer:
this is for homework not for test or quizzes
Mike jogged a total of A 5 and 1/3 times 1/6
your welcome.
I cannot reach a meaningful solution from the given information. To prove that S was always true, you would have to prove that N was always false. To prove that N was always false you would have to prove that L was always false. For the statement (L ^ T) -> K to be true, you only need K to be true, so L can be either true or false.
Therefore, because of the aforementioned knowledge, I do not believe that you can prove S to be true.
First picture)
I: 5x+2y=-4
II: -3x+2y=12
add I+(-1*II):
5x+2y-(-3x+2y)=-4-12
8x=-16
x=-2
insert x=-2 into I:
5*(-2)+2y=-4
-10+2y=-4
2y=6
y=3
(-2,3)
question 6)
I: totalcost=115=3*childs+5*adults
II: 33=adults+childs
33-adults=childs
insert childs into I:
115=3*(33-adults)+5*adults
115=99-3*adults+5*adults
16=2*adults
8=adults
insert adults into II:
33-8=childs
25=childs
so it's the last option
question 7)
a) y<6 and y>2 can also be written as 2<y<6, so solution 3 exist for example
b) y>6 and y>2 can also be written as 2<6<y, so solution 7 exist for example
c) y<6 and y<2 inverse of b: y<2<6, so for example 1
d) y>6 and y<2: y<2<6<y, this is impossible as y can be only either bigger or smaller than 2 or 6
so it's the last option
question 8)
I: x+y=12
II: x-y=6
subtract: I-II:
x+y-(x-y)=12-6
2y=6
y=3
insert y into I:
x+3=12
x=9
(9,3)
question 9)
I: x+y=6
II: x=y+5
if you take the x=y+5 definition of II and substitute it into I:
(y+5)+y=6
which is the second option :)
Answer:
the answer b
Step-by-step explanation:
[surface area]=[area of 4 triangles sides]+[area of square base]
[area of square base]=24*24--------> 576 cm²
[area of one triangles side]
l=slant height
l²=16²+12²-------> l²=400-----------> l=20 cm
[area of one triangles side]=24*20/2--------> 240 cm²
[area of 4 triangles sides]=4*240----------> 960 cm²
[surface area]=[960]+[576]--------> 1536 cm²
