The quadratic equation has two solutions if b^2 - 4ac > 0
Given the equation ax^4 + bx^2 + c=0
Substitute into the formula to have:
The equation becomes aP^2 + bP + c = 0
For us to have a unique solution, the discriminant b^2 - 4ac must be greater than zero. Hence the quadratic equation has two solutions if b^2 - 4ac > 0
learn more on discriminant here; brainly.com/question/1537997
4 + 5 = 9
180 ÷ 9 = 20
Sand = 20×4 = 80
Cement = 20×5 = 100
Standard form : Ax + By = C
y - 4 = 0
y = 4...this is a horizontal line and a horizontal line has a 0 slope
so in standard form, ur equation is : 0x + y = 4
Answer:
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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
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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
