Let First Sphere be the Original Sphere
its Radius be : r
We know that Surface Area of the Sphere is : 4π × (radius)²
⇒ Surface Area of the Original Sphere = 4πr²
Given : The Radius of Original Sphere is Doubled
Let the Sphere whose Radius is Doubled be New Sphere
⇒ Surface of the New Sphere = 4π × (2r)² = 4π × 4 × r²
But we know that : 4πr² is the Surface Area of Original Sphere
⇒ Surface of the New Sphere = 4 × Original Sphere
⇒ If the Radius the Sphere is Doubled, the Surface Area would be enlarged by factor : 4
So you take $37.00 minus 35% and that gets $ 12.95
F=ir^t
139=134r^10
139/134=r^10
r=(139/134)^(1/10) then:
f=134(139/134)^(t/10) so in 2014, t=24 so
f=134(139/134)^(2.4)
f≈146 million (to nearest million)
Some will say that you have to use the exponential function, but it really gives you the same answer...even for continuous compounding :)...
A=Pe^(kt)
139=134e^(10k)
139/134=e^(10k)
ln(139/134)=10k
k=ln(139/134)/10 so
A=134e^(t*ln(139/134)/10) when t=24
A=134e^(2.4*ln(139/134))
A≈146 million (to nearest million)
The only real reason or advantage to using A=Pe^(kt) is when you start getting into differential equations...
The answer is below
Step-by-step explanation:
Two triangles are said to be congruent if for the two triangles all the three sides are equal and all the three angles are equal to each other.
Given that ΔPQR is congruent to ΔSTU, hence all corresponding sides and corresponding angles are equal to each other.
∠P = ∠S = 59°
∠S + ∠T + ∠U = 180° (sum of angles in a triangle)
59 + 75 + ∠U = 180°
134 + ∠U = 180°
∠U = 46°
∠U = ∠R = 46°
∠Q = ∠T = 75°
Also, for the sides:
ST = PQ = 7 m
SU = PR = 8 m
Yes, real numbers are always rational numbers
