2a.
Since BC = 10m and M is the midpoint, we can find CM by : 10÷2 = 5m
So, CM = 5m
2b. Now, you have the base ( 5m) & hypotenuse (13m). Pythagoras' Theorem:
Where c is the hypotenuse, a & b are sides.
b here refers to the height.
2c.
Area of a right-angled triangle =
So,
Find slope first:
m =y2-y1/x2-x1
m = 1--2/2--2
m = 3/4
Select a point, insert your slope, & put it into point slope form:
y-y1 =m(x-x1)
Your final answers are:
y-1=3/4(x-2)
OR
y+2 =3/4(x+2)
2 circles and for the part in the middle if you look at it as a net it
will be a rectangle just rolled up. That is called a lateral surface or a
lateral face
Answer:
x=7
Step-by-step explanation:
8x +3 = 59
8x= 56
/8
x = 7
Standard reduction of order procedure: suppose there is a second solution of the form
, which has derivatives
Substitute these terms into the ODE:
and replacing
, we have an ODE linear in
:
Divide both sides by
, giving
and noting that the left hand side is a derivative of a product, namely
![\dfrac{\mathrm d}{\mathrm dx}[wx]=0](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Bwx%5D%3D0)
we can then integrate both sides to obtain
Solve for
:
Now
where the second term is already accounted for by
, which means
, and the above is the general solution for the ODE.
