Uhhh , what are we suppose to answer here ?
m=-(5/4)
From left to right, (1,3) is first and then comes (5,-2). Always remember when finding slopes without equations, the rule is RISE over RUN, to the numerator and denominator, respectively.
The y value of the second coordinates becomes negative which is unlike the y value in the first coordinates, which means our slope is downward, meaning it has a negative sign in front.
In every slope, there’s a numerator, being the rise, and a denominator, being the run.
To find the rise, we must look at the y values. Starting at 3 going to -2 has a space of 5 units, making that our numerator.
To find the run, the first x value is 1 and the second is 5, making a space of 4, which is out denominator.
With these two numbers and the negative sign, we get -(5/4) as our slope.
I believe is C because the formula is
Y = mx + b
Since your m value is -6/5 and your x value is -3, and y value is -7
Then your answer would be
Y - 7 = -6/5 (x - 3)
Answer:
x = 10, x = 124
Step-by-step explanation:
(5)
Since Δ JKL is isosceles then the base angles are congruent, that is
∠ LJK = ∠ LJG = 80° , thus
∠ JLK = 180 - (80 + 80) = 180 - 160 = 20
x = 20 ÷ 2 = 10
(6)
Since Δ PQR is equilateral, then each of its angles = 60°
∠QRS = 60° - 32° = 28°
Since Δ SQR is isosceles then base angles are congruent, then
∠ SQR = ∠ QRS = 28 , thus
x = 180 - (28 + 28) = 180 - 56 = 124
By the chain rule,
which follows from 
.
is then a function of 
; denote this function by 
. Then by the product rule,
![\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1x\dfrac{\mathrm dy}{\mathrm dt}\right]=-\dfrac1{x^2}\dfrac{\mathrm dy}{\mathrm dt}+\dfrac1x\dfrac{\mathrm df}{\mathrm dx}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%5E2y%7D%7B%5Cmathrm%20dx%5E2%7D%3D%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cleft%5B%5Cdfrac1x%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dt%7D%5Cright%5D%3D-%5Cdfrac1%7Bx%5E2%7D%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dt%7D%2B%5Cdfrac1x%5Cdfrac%7B%5Cmathrm%20df%7D%7B%5Cmathrm%20dx%7D)
and by the chain rule,
so that
Then the ODE in terms of 
is
The characteristic equation
has two roots at 
and 
, so the characteristic solution is
Solving in terms of gives
