Let's represent the two numbers by x and y. Then xy=60. The smaller number here is x=y-7.
Then (y-7)y=60, or y^2 - 7y - 60 = 0. Use the quadratic formula to (1) determine whether y has real values and (2) to determine those values if they are real:
discriminant = b^2 - 4ac; here the discriminant is (-7)^2 - 4(1)(-60) = 191. Because the discriminant is positive, this equation has two real, unequal roots, which are
-(-7) + sqrt(191)
y = -------------------------
-2(1)
and
-(-7) - sqrt(191)
y = ------------------------- = 3.41 (approximately)
-2(1)
Unfortunately, this doesn't make sense, since the LCM of two numbers is generally an integer.
Try thinking this way: If the LCM is 60, then xy = 60. What would happen if x=5 and y=12? Is xy = 60? Yes. Is 5 seven less than 12? Yes.
No area is shown. So befoe sam divided it,
the area of tge original rectangle was . . .
8 x 0 = zero .
Answer:
The general equation of a circle is x2 + y2 - 6x - 16y + 48 = 0
Step-by-step explanation:
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
Answer:
Step-by-step explanation:
A1. C = 104°, b = 16, c = 25
Law of Sines: B = arcsin[b·sinC/c} ≅ 38.4°
A = 180-C-B = 37.6°
Law of Sines: a = c·sinA/sinC ≅ 15.7
A2. B = 56°, b = 17, c = 14
Law of Sines: C = arcsin[c·sinB/b] ≅43.1°
A = 180-B-C = 80.9°
Law of Sines: a = b·sinA/sinB ≅ 20.2
B1. B = 116°, a = 11, c = 15
Law of Cosines: b = √(a² + c² - 2ac·cosB) = 22.2
A = arccos{(b²+c²-a²)/(2bc) ≅26.5°
C = 180-A-B = 37.5°
B2. a=18, b=29, c=30
Law of Cosines: A = arccos{(b²+c²-a²)/(2bc) ≅ 35.5°
Law of Cosines: B = arccos[(a²+c²-b²)/(2ac) = 69.2°
C = 180-A-B = 75.3°
