E is exponents, D is division, A is addition
Answer:
Follows are the solution to the given point:
Step-by-step explanation:
In point a:
¬∃y∃xP (x, y)
∀x∀y(>P(x,y))
In point b:
¬∀x∃yP (x, y)
∃x∀y ¬P(x,y)
In point c:
¬∃y(Q(y) ∧ ∀x¬R(x, y))
∀y(> Q(y) V ∀ ¬ (¬R(x,y)))
∀y(¬Q(Y)) V ∃xR(x,y) )
In point d:
¬∃y(∃xR(x, y) ∨ ∀xS(x, y))
∀y(∀x>R(x,y))
∃x>s(x,y))
In point e:
¬∃y(∀x∃zT (x, y, z) ∨ ∃x∀zU (x, y, z))
∀y(∃x ∀z)>T(x,y,z)
∀x ∃z> V (x,y,z))
For this item, I will assume that we are required to give the area of the dilated triangle. By dilated we mean to say that the dimensions of the second triangle is 6 times that of the first. We square 6 and them multiply this to the area of the original triangle to get the area of the second. That is,
area of second triangle = (2/3 cm²)(6²) = 24 cm²
Thus, the area of the new triangle is equal to 24 cm².
we have

we know that
The radicand must be greater than or equal to zero
so

the domain is is the interval--------> [-2.25,∞)
therefore
<u>the answer is</u>

Q55.
UTA+ATS=UTS
x+18+120=12x+17
x+138=12x+17
12x-x=138-17
11x=121
x=121/11
x=11
mUTA=x+18
put the value of x
11+18
mUTA=29
Q57.
AQP+RQA=RQP
9x+2+75=1+28x
9x+77=1+28x
28x-9x=77-1
19x=76
x=76/19
x=4
mRQP=1+28x
put the value of x
1+28(4)
1+112
mRQP=113