(3x−1)+x2−4 when x=3 how do i solve it
1 answer:
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Answer:
6m
Step-by-step explanation:
the side of first square = 3 m
area of first square = 3^2= 9 m^2
let the side of second square = x m
since diagonal of first square = x m as per the question
in a square each angle is 90 degree
therefore by applying pythagorus theorem ,
diagonal ^2 = side^2 + side ^2
diagonal^2 = 3^2+3^2=9+9
DIAGONAL= square root of 18
diagonal= 3 m = side of second square
therefore in second square ,
one angle is 90 degree , therefore by applying pythagorue theorem
diaqonal^2 = side ^2 + side ^2 = (3)^2 + ( 3
)^2
diagonal ^2 = 18 + 18
diagonal = square root of 36
therefore diagonal of second square is 6m
hope it helps you... please mark me as the brainliest..
Answer:
Step-by-step explanation:
Let f(x) be the polynomial function of minimum degree with real coefficients whose zeros are 5, -3, and -1 + 3i be f(x).
By the complex conjugate property of polynomials, -1-3i is also a root of this polynomial.
Therefore the polynomial in factored form is
We expand to get:
We expand further to get:\