Answer:
Hi there!
Your answer is:
h(5y) = 5y-9
Step-by-step explanation:
h(5y) means we substitute 5y for p! This completely gets rid of the p!
Hope this helps
Foil it out:
(7x)(4x)+(7x)(9)+(-8)(4x)+(-8)(9)
28x^2+63x-32x-72
28x^2+31x-72
Answer:
(a) x=5
(b) x= -1 1/3
(c) x= 1
Step-by-step explanation:
(a) (b) (c)
2(x-3)=4 −3(2x+1)=5 a(bx+c)=d
x−3=4/2 2x+1= − 5/3 d^1−1
x−3=2 2x= -5/3-3/3 d^0
x=2+3 2x= -5-3/3 x=1
x=5 2x= -8/3*2
x= -8/6
x= -4/3
Throughout all of these steps I'm only going to alter the left hand side (LHS). I am NOT going to change the right hand side (RHS) at all.
Before I change the LHS of the original equation, let's focus on the given identity
cot^2(x) + 1 = csc^2(x)
Since we know it's an identity, we can subtract 1 from both sides and the identity would still hold true
cot^2(x) + 1 = csc^2(x)
cot^2(x) + 1-1 = csc^2(x)-1
cot^2(x) + 0 = csc^2(x)-1
cot^2(x) = csc^2(x)-1
So we'll use the identity cot^2(x) = csc^2(x)-1
---------------------------------------------
Now onto the main equation given
cot^2(x) + csc^2(x) = 2csc^2(x) - 1
cot^2(x) + csc^2(x) = 2csc^2(x) - 1 .... note the term in bold
csc^2(x)-1 + csc^2(x) = 2csc^2(x) - 1 .... note the terms in bold
[ csc^2(x) + csc^2(x) ] - 1 = 2csc^2(x) - 1
[ 2csc^2(x) ] - 1 = 2csc^2(x) - 1
2csc^2(x) - 1 = 2csc^2(x) - 1
The bold terms indicate how the replacements occur.
So the original equation has been proven to be an identity because the LHS has been altered to transform into the RHS
Answer:
9 and 7
Step-by-step explanation:
multiply 7 and 1 after that it will 7 and again adding 7 + 2 so it will be 9 so the answer will be 9 and 7
