B because
Solve for the first variable in one of the equations, then substitute the result into the other equation.
To answer you get the same denominator so multiply both denominators with each other and do the same with the top so 25 will be at both bottoms and well 15/25 and 8/25. Then solve
Answer:
x²+6x+9=0
x²+3x+3x+9=0
x(x+3)+3(x+3)=0
(x+3)(x+3)=0
either
x+3=0
x=-3
or
x+3=0
x=-3
Step-by-step explanation:
next method
x²+6x+9=0
x²+2×x×3+3²=0
it is in formula of (x+y)²
(x+3)²=0
x+3=√0
x+3=0
x=-3
My answer -
<span>1. Use symbols (not words) to express quotient
2. Use exponent symbol (^) to denote exponents
3. Just write out question number, question, and choices. No need for
extra information (such as points). Also, don't leave blank lines
between choices. This extraneous that we don't need just makes your
whole question very very long, and means a lot of scrolling on our part.
4. You should only post 2 or 3 questions at a time.
1) (6x^3 − 18x^2 − 12x) / (−6x) = −x^2 + 3x + 2 ----> so much simpler to read !
2) (d^7 g^13) / (d^2 g^7) = d^(7−2) g^(13−7) = d^5 g^6 ----> much easier to read !
3) (4x − 6)^2 = 16x^2 − 24x − 24x + 36 = 16x^2 − 48x + 36
4) (x^2 / y^5)^4 = (x^2)^4 / (y^5)^4 = x^8 / y^20
5) (3x + 5y)(4x − 3y) = 12x^2 − 9xy + 20xy − 15y^2 = 12x^2 + 11xy − 15y^2
6) (3x^3y^4z^4)(2x^3y^4z^2) = (3*2) x^(3+3) y^(4+4) z^(4+2) = 6 x^6 y^8 z^6
7) 5x + 3x^4 − 7x^3 ----> Fourth degree trinomial
8) (5x^3 − 5x − 8) + (2x^3 + 4x + 2) = 7x^3 − x − 6
9) (x − 1) + (2x + 5) − (x + 3) = x + 1
10) (−4g^8h^5k^2)0(hk^2)^2 = 0 (anything multiplied by 0 = 0)
or.. (−4g^8h^5k^2)^0(hk^2)^2 = 1 (h^2 (k^2)^2) = h^2 k^4
Last question shows why it is so important to use proper symbols (such
as ^ to indicate exponents). Without such symbols, I could not tell if
the 0 was an actual number and part of multiplication, of if 0 was an
exponent of the expression preceding it.
P.S
Glad to help you have an AWESOME!!! day :)
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