The sum of the cost of dinner d and a rip of $12 tip divided equally by 3 people

Devon invests in an account for 10 years that pays 2.15% compound interest annually. He uses the expression P(1+r)t to find the

zheka24 [161]

Answer:

FV= $4,948.16

Step-by-step explanation:

Giving the following information:

Number of periods (n)= 10

Interest rate (i)= 2.15% compound interest annually

Present value (PV)= 4,000

To calculate the future value, we need to use the following formula:

FV= PV*(1+i)^n

FV= 4,000*(1.0215^10)

FV= $4,948.16

7 0

3 years ago

X = ????????? Geometry

bazaltina [42]

Answer:

$\boxed{x = 6}$

Step-by-step explanation:

Using Chord-Chord Power theorem:

=> (2)(x) = (4)(3)

=> 2x = 12

Dividing both sides by 2

=> x = 6

7 0

3 years ago

What is the mean of Variable A?

hammer [34]

The answer is 5.6 because 2+2+4+5+5+6+7+7+8+10=56
and 56/10=5.6
To find the mean, add all the values together and divide by the number of values in the set. The result is your mean!
Hope I helped:)

6 0

3 years ago

You give tours on the Great Lakes and talk about points of interest. A lighthouse typically has 2 beacons that rotate together—b

tatiyna

Since the beacons rotate together, the angular speed is constant.

From the time you see the first beacon, you have to wait for the lighthouse to make 120°, i.e. 1/3 of a whole turn.

So, if T is the time it takes for the lighthouse to make a whole turn, the interval between the first and the second beacon is T/3.

The interval between the second and third beacon is the amount of time it takes to the lighthouse to make the remaining 2/3 of a turn, so that you'll see the first beacon again. So, this time is 2T/3

So, the ratio is

$\dfrac{2T}{3}\div\dfrac{T}{3}=\dfrac{2T}{3}\cdot\dfrac{3}{T}=2$

So, the time between the second and third beacon is twice as much the time between the first two.

5 0

3 years ago

How would I find the integral of <img src="https://tex.z-dn.net/?f=%5Cint%5Cfrac%7Btdt%7D%7Bt%5E4%2B2%7D" id="TexFormula1" title

kotegsom [21]

Let $t=\sqrt y$ , so that $t^2=y$ , $t^4=y^2$ , and $\mathrm dt=\dfrac{\mathrm dy}{2\sqrt y}$ . Then

$\displaystyle\int\frac t{t^4+2}\,\mathrm dt=\int\frac{\sqrt y}{2\sqrt y(y^2+2)}\,\mathrm dy=\frac12\int\frac{\mathrm dy}{y^2+2}$

Now let $y=\sqrt2\tan z$ , so that $\mathrm dy=\sqrt2\sec^2z\,\mathrm dz$ . Then

$\displaystyle\frac12\int\frac{\mathrm dy}{y^2+2}=\frac12\int\frac{\sqrt2\sec^2z}{(\sqrt2\tan z)+2}\,\mathrm dz=\frac{\sqrt2}4\int\frac{\sec^2z}{\tan^2z+1}\,\mathrm dz=\frac1{2\sqrt2}\int\mathrm dz=\dfrac1{2\sqrt2}z+C$

Transform back to $y$ to get

$\dfrac1{2\sqrt2}\arctan\left(\dfrac y{\sqrt2}\right)+C$

and again to get back a result in terms of $t$ .

$\dfrac1{2\sqrt2}\arctan\left(\dfrac{t^2}{\sqrt2}\right)+C$

3 0

3 years ago