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3 years ago

Please help me with both questions :(

Mathematics

1 answer:

MrRissso [65]3 years ago

7 0

Answer:

2.) $520

3.) less

Step-by-step explanation:

I'm going to assume that the interst is compoudning and is convertable once a year

The compound interest formula for interest compounding only once a year is as follows

$AV=PV(1+i)^n$

plug in the numbers and get

$500(1+.04)^{1}=500*1.04=520$

3.) If the interest rate is lower at the credit union he would obvioulsy be paying less (assuming that this interest rate is convertable annually as well)

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Answer:

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3 years ago

Sand is leaking through a small hole at the bottom of a conical funnel at the rate of 12cm3/s . if the radius of the funnel is 8

Bezzdna [24]

Answer:

h ≅ 16.29 cm

$\mathbf{\dfrac{dh}{dt}= -0.1809 \ cm/s}$

Step-by-step explanation:

From the conical funnel;

Let the adjacent line at the base of the cone be radius (r) and the opposite ( vertical length) be the height (h)

Then;

$tan \theta= \dfrac{r}{h}$

$tan \theta= \dfrac{8}{16}$

$tan \theta= \dfrac{1}{2}$

∴

$\dfrac{r}{h} = \dfrac{1}{2}$

$r = \dfrac{h}{2}$

The volume (V) of the cone is expressed as:

$V = \dfrac{1}{3}\pi r ^2 h$

$V = \dfrac{1}{3}\pi (\dfrac{h}{2})^2 h$

$V = \dfrac{\pi h^3}{3\times 4}$

$\dfrac{dv}{dt} = \dfrac{3 \pi h^2}{3 \times 4} \dfrac{dh}{dt}$

$\dfrac{dv}{dt} = \dfrac{ \pi h^2}{4} \dfrac{dh}{dt}$

Given that:

$\dfrac{dv}{dt} = 12 \pi$

Then:

$-12 \pi = \dfrac{ \pi h^2}{4} \dfrac{dh}{dt}$

$- \int 48 \ dt = \int h^2 \ dh$

$\dfrac{h^3}{3}= -48 t + c$

$At \ t = 0 \ ; h = 0$

∴

$\dfrac{0^3}{3} = -48(0) + c$

c = 0

So;

$\dfrac{h^3}{3}= -48 t + 0$

$\dfrac{h^3}{3}= -48 t$

t = 30

$\implies h = (-48(30))^{1/3}$

h = 16.2865

h ≅ 16.29 cm

Thus;

from $-12 \pi = \dfrac{ \pi h^2}{4} \dfrac{dh}{dt}$

$\dfrac{-12 \pi} { \dfrac{ \pi h^2}{4} } =\dfrac{dh}{dt}$

$\dfrac{dh}{dt} = \dfrac{-12 \pi} { \dfrac{ \pi h^2}{4} }$

$\dfrac{dh}{dt}=\dfrac{-12 \times 4} { {16.29^2} }$

$\mathbf{\dfrac{dh}{dt}= -0.1809 \ cm/s}$

7 0

3 years ago