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3 years ago

14

sitting on park bench, you see a swing that is 60 feet away and a slide that is 80 feet away. The angle between them is 30 degre

es. What is the approximate distance between the swing and the slide?

2 answers:

Snowcat [4.5K]3 years ago

4 0

I would use the Pythagorean theorem

slavikrds [6]3 years ago

4 0

Answer:41 feet

Step-by-step explanation:

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Answer:

$\frac{5}{6}$

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ExtremeBDS [4]

Answer:

95

Step-by-step explanation:

you first find out the angles mn and mt.

Since ns and ms are the same, we know ms is 35 degrees.

mt is 50, so we subtract 50 and 35 from 180, which is the degree of m.

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The function in Exercise represents the rate of flow of money in dollars per year. Assume a 10-year period at 8% compounded cont

anzhelika [568]

Answer:

a) The present value is 688.64 $

b) The accumulated amount is 1532.60 $

Step-by-step explanation:

<u>a)</u><u> The preset value equation is given by this formula:</u>

$P=\int^{T}_{0}f(t)e^{-rt}dt$

where:

T is the period in years (T = 10 years)
r is the annual interest rate (r=0.08)

So we have:

$P=\int^{T}_{0}(0.01t+100)e^{-rt}dt$

Now we just need to solve this integral.

$P=\int^{T}_{0}0.01te^{-rt}dt+\int^{T}_{0}100e^{-rt}dt$

$P=e^{-0.08t}(-1.56-0.13t)|^{10}_{0}+1250e^{-0.08t}|^{10}_{0}$

$P=0.30+688.34=688.64 $$

The present value is 688.64 $

<u>b)</u><u> The accumulated amount of money flow formula is:</u>

$A=e^{r\tau}\int^{T}_{0}f(t)e^{-rt}dt$

We have the same equation but whit a term that depends of τ, in our case it is 10.

So we have:

$A=e^{r\tau}\int^{T}_{0}(0.01t+100)e^{-rt}dt=e^{0.08\cdot 10}P$

$A=e^{0.08\cdot 10}688.64=1532.60 $$

The accumulated amount is 1532.60 $

Have a nice day!

6 0

3 years ago

A liquid at temperature 7575F is placed in an oven at temperature 450450. The temperature of the liquid increases at a rate 77 t

hichkok12 [17]

Answer: $\dfrac{dT(t)}{dt}=77\times (450-T)$

Step-by-step explanation:

Given

The temperature of the liquid is $75^{\circ}F$ placed in an oven with temperature of $450^{\circ}F$ .

Initially difference in temperature of the two

$\Delta T=450-75\\\Rightarrow \Delta T=375^{\circ}F$

According to the question

$\Rightarrow \dfrac{dT(t)}{dt}=77\cdot \Delta T\\\\\Rightarrow \dfrac{dT(t)}{dt}=77\times (450-T)\quad [\text{T=75}^{\circ}F\ \text{at t=0}]$

5 0

3 years ago