There is only one solution since x=0 x cant be many things since the only thing that would make this true is 0
Answer:
yes
Step-by-step explanation:
I think sorry if I am wrong
This may not be very pretty.
(2x + 5) * (2x + 5) expanded is (use foil)
F = 4x^2
O = 2x * 5 = 10x
I = 2x + 5 = 10x
L = 25
Total = 4x^2 + 20x + 25
(x + 3)(x + 3) = x^2 + 6x + 9 by the same method
(2x + 7)(2x + 7) = 4x^2 + 28x + 49 Same method.
4x^2 + 20x + 25 + x^2 + 6x + 9 = 4x^2 + 28x + 49
5x^2 + 26x + 34 = 4x^2 + 28x + 49 Collect everything on the left.
x^2 - 2x - 15 = 0
(x - 5)(x + 3) = 0
x + 3 will have no meaning.
x - 5 = 0
x = 5
2x + 5 = 2*5 + 5 = 15
x + 3 = 5 + 3 = 8
2x+ 7 = 2*5 +7 = 17
Check
15^2 + 8^2 = 17^2
225 + 64 = ? 289
289 = 289
a = 15 <<<<< answer
b = 8 <<<<< answer
c = 17 <<<<< answer
Answer:
y = 
Domain = [3, infinity)
Step-by-step explanation:
x = 3y^2 + 3
3y^2 = x - 3
y^2 = (x-3)/3
y =
Using pythagorean identities,tan(sin^-1(-5/13))= tan(arcsin(-5/13))
Let A = arcsin(-5/13) = -arcsin(5/13)
Thus, sin A = -5/13 and cos A = sqrt(1 - (5/13)^2) = 12/13.= tan(A)= sin(A) / cos(A)= -5/13 / (12/13)= -5/12
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