Answer:
the distance is 16
Step-by-step explanation:
Hi there!
We are given point A (-4,-13) and point B (-4,3). We need to find the distance between those two points
the distance formula is given as
where (
,
) and (
,
) are points
we are given 2 points, which is what we need for the formula. However, let's label the values of the points to avoid any confusion
=-4
=-13
=-4
=3
now substitute those values into the formula. Remember: the formula uses SUBTRACTION.
![\sqrt{(-4--4)^2+(3--13)^2}](https://tex.z-dn.net/?f=%5Csqrt%7B%28-4--4%29%5E2%2B%283--13%29%5E2%7D)
simplify
![\sqrt{(-4+4)^2+(3+13)^2}](https://tex.z-dn.net/?f=%5Csqrt%7B%28-4%2B4%29%5E2%2B%283%2B13%29%5E2%7D)
now add the values inside the parenthesis that are under the radical
![\sqrt{(0)^2+(16)^2}](https://tex.z-dn.net/?f=%5Csqrt%7B%280%29%5E2%2B%2816%29%5E2%7D)
raise everything under the radical to the second power
![\sqrt{0+256}](https://tex.z-dn.net/?f=%5Csqrt%7B0%2B256%7D)
add under the radical
![\sqrt{256}](https://tex.z-dn.net/?f=%5Csqrt%7B256%7D)
now take the square root of 256
=16
so the distance between point A and point B is <u>16</u>
Hope this helps! :)
Half of R is shown as 20 degrees, so the other half of r is also 20 degrees.
A tangent line forms a right angle, so the tangent at P is 90 degrees.
X would be 180 - 90 - 20 = 70 degrees.
X = 70 degrees.
Answer:
6 ketchup / 1 mustard
Step-by-step explanation:
3 ketchup = 0.5 mustard
6 ketchup per 1 mustard
Answer:
![A = \frac{P}{r}\left( e^{rt} -1 \right)](https://tex.z-dn.net/?f=A%20%3D%20%5Cfrac%7BP%7D%7Br%7D%5Cleft%28%20e%5E%7Brt%7D%20-1%20%5Cright%29)
Step-by-step explanation:
This is <em>a separable differential equation</em>. Rearranging terms in the equation gives
![\frac{dA}{rA+P} = dt](https://tex.z-dn.net/?f=%5Cfrac%7BdA%7D%7BrA%2BP%7D%20%3D%20dt)
Integration on both sides gives
![\int \frac{dA}{rA+P} = \int dt](https://tex.z-dn.net/?f=%5Cint%20%5Cfrac%7BdA%7D%7BrA%2BP%7D%20%3D%20%5Cint%20%20dt)
where
is a constant of integration.
The steps for solving the integral on the right hand side are presented below.
![\int \frac{dA}{rA+P} = \begin{vmatrix} rA+P = m \implies rdA = dm\end{vmatrix} \\\\\phantom{\int \frac{dA}{rA+P} } = \int \frac{1}{m} \frac{1}{r} \, dm \\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \int \frac{1}{m} \, dm\\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |m| + c \\\\&\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |rA+P| +c](https://tex.z-dn.net/?f=%5Cint%20%5Cfrac%7BdA%7D%7BrA%2BP%7D%20%3D%20%5Cbegin%7Bvmatrix%7D%20rA%2BP%20%3D%20m%20%5Cimplies%20rdA%20%3D%20dm%5Cend%7Bvmatrix%7D%20%5C%5C%5C%5C%5Cphantom%7B%5Cint%20%5Cfrac%7BdA%7D%7BrA%2BP%7D%20%7D%20%3D%20%5Cint%20%5Cfrac%7B1%7D%7Bm%7D%20%5Cfrac%7B1%7D%7Br%7D%20%5C%2C%20dm%20%5C%5C%5C%5C%5Cphantom%7B%5Cint%20%5Cfrac%7BdA%7D%7BrA%2BP%7D%20%7D%20%3D%20%5Cfrac%7B1%7D%7Br%7D%20%5Cint%20%5Cfrac%7B1%7D%7Bm%7D%20%5C%2C%20dm%5C%5C%5C%5C%5Cphantom%7B%5Cint%20%5Cfrac%7BdA%7D%7BrA%2BP%7D%20%7D%20%3D%20%5Cfrac%7B1%7D%7Br%7D%20%5Cln%20%7Cm%7C%20%2B%20c%20%5C%5C%5C%5C%26%5Cphantom%7B%5Cint%20%5Cfrac%7BdA%7D%7BrA%2BP%7D%20%7D%20%3D%20%5Cfrac%7B1%7D%7Br%7D%20%5Cln%20%7CrA%2BP%7C%20%2Bc)
Therefore,
![\frac{1}{r} \ln |rA+P| = t+c](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Br%7D%20%5Cln%20%7CrA%2BP%7C%20%3D%20t%2Bc)
Multiply both sides by ![r.](https://tex.z-dn.net/?f=r.)
![\ln |rA+P| = rt+c_1, \quad c_1 := rc](https://tex.z-dn.net/?f=%5Cln%20%7CrA%2BP%7C%20%3D%20rt%2Bc_1%2C%20%5Cquad%20c_1%20%3A%3D%20rc)
By taking exponents, we obtain
![e^{\ln |rA+P|} = e^{rt+c_1} \implies |rA+P| = e^{rt} \cdot e^{c_1} rA+P = Ce^{rt}, \quad C:= \pm e^{c_1}](https://tex.z-dn.net/?f=e%5E%7B%5Cln%20%7CrA%2BP%7C%7D%20%3D%20e%5E%7Brt%2Bc_1%7D%20%5Cimplies%20%20%7CrA%2BP%7C%20%3D%20e%5E%7Brt%7D%20%5Ccdot%20e%5E%7Bc_1%7D%20rA%2BP%20%3D%20Ce%5E%7Brt%7D%2C%20%5Cquad%20C%3A%3D%20%5Cpm%20e%5E%7Bc_1%7D)
Isolate
.
![rA+P = Ce^{rt} \implies rA = Ce^{rt} - P \implies A = \frac{C}{r}e^{rt} - \frac{P}{r}](https://tex.z-dn.net/?f=rA%2BP%20%3D%20Ce%5E%7Brt%7D%20%5Cimplies%20rA%20%3D%20Ce%5E%7Brt%7D%20-%20P%20%5Cimplies%20A%20%3D%20%5Cfrac%7BC%7D%7Br%7De%5E%7Brt%7D%20-%20%5Cfrac%7BP%7D%7Br%7D)
Since
when
, we obtain an initial condition
.
We can use it to find the numeric value of the constant
.
Substituting
for
and
in the equation gives
![0 = \frac{C}{r}e^{0} - \frac{P}{r} \implies \frac{P}{r} = \frac{C}{r} \implies C=P](https://tex.z-dn.net/?f=0%20%3D%20%5Cfrac%7BC%7D%7Br%7De%5E%7B0%7D%20-%20%5Cfrac%7BP%7D%7Br%7D%20%5Cimplies%20%5Cfrac%7BP%7D%7Br%7D%20%3D%20%5Cfrac%7BC%7D%7Br%7D%20%5Cimplies%20C%3DP)
Therefore, the solution of the given differential equation is
![A = \frac{P}{r}e^{rt} - \frac{P}{r} = \frac{P}{r}\left( e^{rt} -1 \right)](https://tex.z-dn.net/?f=A%20%3D%20%5Cfrac%7BP%7D%7Br%7De%5E%7Brt%7D%20-%20%5Cfrac%7BP%7D%7Br%7D%20%3D%20%5Cfrac%7BP%7D%7Br%7D%5Cleft%28%20e%5E%7Brt%7D%20-1%20%5Cright%29)