example: the line is y = ax + b
The line go through 2 points ( -4 ; 0 ) and ( 0 ; -2 )
So:
- -2 = 0a + b => b = -2
- 0 = -4a + (-2) => a = 2/-4 = -1/2
The line is y = -1/2x - 2 ( i'm not sure)
Answer: a
Ok done. Thank to me :>
<span>It is because even numbers always have a factor of two, and therefore, larger composite even numbers will have factors of two and other even numbers based around two, such as 4, 8, 16, 32, and so on. On the other hand, numbers which are odd can have factors of 3, 5, and 7 for example, and their numbers based around them(3, 9, 27; 5, 10, 15; 7, 49, 343; and so on). If we look into it, notice how for odd numbers the space between the numbers based around 3, 5, and 7 are increasingly further apart. This is the reason why less large odd integers to have numerous factors. It is because odd numbers cannot have the prime factor 2, this will reduce their factor number. And is is also because even numbers are already divided by 2, this will give them more factors over the odd numbers.</span>
Answer:
y = ⁹/₄x -9
Step-by-step explanation:
Brainliest Please!!
Step-by-step explanation:
∛9 × ∛-24
∛-216
∛(-6)³
-6
√(8y⁵) √(40y²)
√(320y⁷)
√(64y⁶ × 5y)
8y³√(5y)
Answer:
-6re−r [sin(6θ) - cos(6θ)]
Step-by-step explanation:
the Jacobian is ∂(x, y) /∂(r, θ) = δx/δθ × δy/δr - δx/δr × δy/δθ
x = e−r sin(6θ), y = er cos(6θ)
δx/δθ = -6rcos(6θ)e−r sin(6θ), δx/δr = -sin(6θ)e−r sin(6θ)
δy/δθ = -6rsin(6θ)er cos(6θ), δy/δr = cos(6θ)er cos(6θ)
∂(x, y) /∂(r, θ) = δx/δθ × δy/δr - δx/δr × δy/δθ
= -6rcos(6θ)e−r sin(6θ) × cos(6θ)er cos(6θ) - [-sin(6θ)e−r sin(6θ) × -6rsin(6θ)er cos(6θ)]
= -6rcos²(6θ)e−r (sin(6θ) - cos(6θ)) - 6rsin²(6θ)e−r (sin(6θ) - cos(6θ))
= -6re−r (sin(6θ) - cos(6θ)) [cos²(6θ) + sin²(6θ)]
= -6re−r [sin(6θ) - cos(6θ)] since [cos²(6θ) + sin²(6θ)] = 1
