Answer: ∆V for r = 10.1 to 10ft
∆V = 40πft^3 = 125.7ft^3
Approximate the change in the volume of a sphere When r changes from 10 ft to 10.1 ft, ΔV=_________
[v(r)=4/3Ï€r^3].
Step-by-step explanation:
Volume of a sphere is given by;
V = 4/3πr^3
Where r is the radius.
Change in Volume with respect to change in radius of a sphere is given by;
dV/dr = 4πr^2
V'(r) = 4πr^2
V'(10) = 400π
V'(10.1) - V'(10) ~= 0.1(400π) = 40π
Therefore change in Volume from r = 10 to 10.1 is
= 40πft^3
Of by direct substitution
∆V = 4/3π(R^3 - r^3)
Where R = 10.1ft and r = 10ft
∆V = 4/3π(10.1^3 - 10^3)
∆V = 40.4π ~= 40πft^3
And for R = 30ft to r = 10.1ft
∆V = 4/3π(30^3 - 10.1^3)
∆V = 34626.3πft^3
<em>Answer: </em>
<em>A = $7,350.00</em>
<em></em>
<em>Step-by-step explanation:</em>
<em>Equation:</em>
<em>A = P(1 + rt)</em>
<em>First, converting R percent to r a decimal</em>
<em>r = R/100 = 9%/100 = 0.09 per year.</em>
<em>Putting time into years for simplicity,</em>
<em>30 months / 12 months/year = 2.5 years.</em>
<em></em>
<em></em>
<em>Solving our equation:</em>
<em>A = 6000(1 + (0.09 × 2.5)) = 7350 </em>
<em>A = $7,350.00</em>
<em>The total amount accrued, principal plus interest, from simple interest on a principal of $6,000.00 at a rate of 9% per year for 2.5 years (30 months) is $7,350.00.</em>
<em>* Therefor, the answer is $7,350.00.</em>
<em>* Hopefully this helps:) Mark me the brainliest:)!!!</em>
We know that
If a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment. (Intersecting Secant-Tangent Theorem)
so
ST²=RT*QT
RT=7 in
QT=23+7-----> 30 in
ST²=7*30-----> 210
ST=√210-----> 14.49 in
the answer is
RT=14.49 in
Answer:
5
Step-by-step explanation:
f(5) means to replace x in the equation with 5 then solve.
f(x) = x^2 +2x
f(5) = 5^2 + 2(5)
f(5) = 25 +10
f(5) = 35
The answer is D)35
