Answer:
x = 42°
Step-by-step explanation:
Linear pair angles or straight line angles.
![\therefore \: \angle JML + \angle LMN = 180 \degree \\ \therefore \: 3x + 54 \degree = 180 \degree \\ \therefore \: 3x = 180 \degree - 54 \degree \\ \therefore \: 3x = 126 \degree \\ \therefore \: x = \frac{126 \degree}{3} \\ \therefore \: x = 42\degree \\](https://tex.z-dn.net/?f=%20%5Ctherefore%20%5C%3A%20%20%5Cangle%20JML%20%20%2B%20%5Cangle%20LMN%20%3D%20180%20%5Cdegree%20%5C%5C%20%20%5Ctherefore%20%5C%3A%203x%20%2B%2054%20%5Cdegree%20%3D%20180%20%5Cdegree%20%5C%5C%20%5Ctherefore%20%5C%3A%203x%20%20%3D%20180%20%5Cdegree%20-%2054%20%5Cdegree%20%5C%5C%20%5Ctherefore%20%5C%3A%203x%20%20%3D%20126%20%5Cdegree%20%20%5C%5C%20%5Ctherefore%20%5C%3A%20x%20%3D%20%20%5Cfrac%7B126%20%5Cdegree%7D%7B3%7D%20%20%5C%5C%20%5Ctherefore%20%5C%3A%20x%20%3D%2042%5Cdegree%20%5C%5C%20)
A quadratic function is a function of the form
![f(x)=ax^2+bx+c](https://tex.z-dn.net/?f=f%28x%29%3Dax%5E2%2Bbx%2Bc)
. The
vertex,
![(h,k)](https://tex.z-dn.net/?f=%28h%2Ck%29)
of a quadratic function is determined by the formula:
![h= \frac{-b}{2a}](https://tex.z-dn.net/?f=h%3D%20%5Cfrac%7B-b%7D%7B2a%7D%20)
and
![k=f(h)](https://tex.z-dn.net/?f=k%3Df%28h%29)
; where
![h](https://tex.z-dn.net/?f=h)
is the
x-coordinate of the vertex and
![k](https://tex.z-dn.net/?f=k)
is the
y-coordinate of the vertex. The value of
![a](https://tex.z-dn.net/?f=a)
determines if the <span>
parabola opens upward or downward; if</span>
![a](https://tex.z-dn.net/?f=a)
is positive, the parabola<span> opens upward and the vertex is the
minimum value, but if </span>
![a](https://tex.z-dn.net/?f=a)
is negative <span>the graph opens downward and the vertex is the
maximum value. Since the quadratic function only has one vertex, it </span><span>could not contain both a minimum vertex and a maximum vertex at the same time.</span>
Answer:
Step-by-step explanation: brainly took away EVERYTHING
Answer : y = x + 1
the solution is in the pic below :)
Answer:
Suppose a population of rodents satisfies the differential equation dP 2 kP dt = . Initially there are P (0 2 ) = rodents, and their number is increasing at the rate of 1 dP dt = rodent per month when there are P = 10 rodents.
How long will it take for this population to grow to a hundred rodents? To a thousand rodents?
Step-by-step explanation:
Use the initial condition when dp/dt = 1, p = 10 to get k;
![\frac{dp}{dt} =kp^2\\\\1=k(10)^2\\\\k=\frac{1}{100}](https://tex.z-dn.net/?f=%5Cfrac%7Bdp%7D%7Bdt%7D%20%3Dkp%5E2%5C%5C%5C%5C1%3Dk%2810%29%5E2%5C%5C%5C%5Ck%3D%5Cfrac%7B1%7D%7B100%7D)
Seperate the differential equation and solve for the constant C.
![\frac{dp}{p^2}=kdt\\\\-\frac{1}{p}=kt+C\\\\\frac{1}{p}=-kt+C\\\\p=-\frac{1}{kt+C} \\\\2=-\frac{1}{0+C}\\\\-\frac{1}{2}=C\\\\p(t)=-\frac{1}{\frac{t}{100}-\frac{1}{2} }\\\\p(t)=-\frac{1}{\frac{2t-100}{200} }\\\\-\frac{200}{2t-100}](https://tex.z-dn.net/?f=%5Cfrac%7Bdp%7D%7Bp%5E2%7D%3Dkdt%5C%5C%5C%5C-%5Cfrac%7B1%7D%7Bp%7D%3Dkt%2BC%5C%5C%5C%5C%5Cfrac%7B1%7D%7Bp%7D%3D-kt%2BC%5C%5C%5C%5Cp%3D-%5Cfrac%7B1%7D%7Bkt%2BC%7D%20%5C%5C%5C%5C2%3D-%5Cfrac%7B1%7D%7B0%2BC%7D%5C%5C%5C%5C-%5Cfrac%7B1%7D%7B2%7D%3DC%5C%5C%5C%5Cp%28t%29%3D-%5Cfrac%7B1%7D%7B%5Cfrac%7Bt%7D%7B100%7D-%5Cfrac%7B1%7D%7B2%7D%20%20%7D%5C%5C%5C%5Cp%28t%29%3D-%5Cfrac%7B1%7D%7B%5Cfrac%7B2t-100%7D%7B200%7D%20%7D%5C%5C%5C%5C-%5Cfrac%7B200%7D%7B2t-100%7D)
You have 100 rodents when:
![100=-\frac{200}{2t-100} \\\\2t-100=-\frac{200}{100} \\\\2t=98\\\\t=49\ months](https://tex.z-dn.net/?f=100%3D-%5Cfrac%7B200%7D%7B2t-100%7D%20%5C%5C%5C%5C2t-100%3D-%5Cfrac%7B200%7D%7B100%7D%20%5C%5C%5C%5C2t%3D98%5C%5C%5C%5Ct%3D49%5C%20months)
You have 1000 rodents when:
![1000=-\frac{200}{2t-100} \\\\2t-100=-\frac{200}{1000} \\\\2t=99.8\\\\t=49.9\ months](https://tex.z-dn.net/?f=1000%3D-%5Cfrac%7B200%7D%7B2t-100%7D%20%5C%5C%5C%5C2t-100%3D-%5Cfrac%7B200%7D%7B1000%7D%20%5C%5C%5C%5C2t%3D99.8%5C%5C%5C%5Ct%3D49.9%5C%20months)