F(x)= x² + 5, is just a parabola shfited upwards by 5 units, so, is a smooth graph and no abrupt edges, so from 0 to 3, is indeed differentiable and continuous. So Rolle's theorem applies, let's check for "c" by simply setting its variable to 0, bear in mind that, looking for "c" in this context, is really just looking for a critical point, since we're just looking where f'(c) = 0, and is a horizontal tangent line.
Answer:
Step-by-step explanation:
(a - b)(a + b) =a² - b²
(3x + 5)(3x -5) =(3x)² - 5² = 9x² - 25
(x +2)(x+2) = (x+ 2)² = x²+ 2*2*x+2² = x² + 4x + 4
(x+2)(3x -5 ) = x*3x - 5*x +2*3x +2*(-5)
= 3x² -5x +6x -10
= 3x² + x - 10


13x² - 20x - 45 = 3*(3x² + x - 10)
13x² - 20x - 45 =9x² + 3x - 30
13x² - 9x² -20x - 3x - 45 + 30 = 0
4x² - 23x - 15 = 0
Answer:
21x-28
Step-by-step explanation:
the distributive property states that a(bx+c) is equal to abx+ac. in this expression, a is 7, b is 3, and c is negative 4. ab is 7*3, which is 21, so abx is 21x. ac is 7*-4, which is -28. combining these two results in 21x-28.
Differentiate:x^2 + y^2 = 4
You get this:2x + 2yy' = 0
bring the x variable on the other side:2yy' = -2x
Now divide 2y to leave y' by itself:y' = -2x/2y
Answer: y' = -x/y
The given relation is not a implicit solution to the given differential equation.
I think that there are 33 left.
I hope this helps you!