Let x=1+.072/12=1.006 the monthly rate of increase of money
Let F be the future value of the cash stream.
F=200×∑k=0,2391.006k=200×(1.006240-1)/.006=106752.46777
the sum is taken from 0 to 239, since the first payment occurs 1 month after the start of the annuity and the last payment...

5x2-(2x-3)2=5x2-(4x2-12x+9)=5x2-4x2+12x-9=x2+12x-9
That's as far as I can go without seeing the choices

y=x2+6x-4=(x+3)2-13
y+13=(x+3)2 vertex at (-3,-13) ,
y-(-13-n)=(x-(-3-m))2 is this parabola shifted m to right, n up. New vertex at (5,-6) means that
-13-n=-6 so n=-7, -3-m=5 so m=-8 and equation is y+6=(x-5)2
m/n=8/7

It is necessary to assume that x1 and x0 are known. This makes the fact that x2=x1-x0 given.
We write
xn+1=xn-xn-1
xn+2=xn+1-xn=xn-xn-1-xn=-xn-1
xn+3=xn+2-xn+1=-xn
xn+4=xn+3-xn+2=-xn+xn-1
xn+5=xn+4-xn+3=xn-1
xn+6=xn+5-xn+4=xn
xn+7=xn+6-xn+5=xn-xn-1=xn+1 ...

Let S be the sum in question. Each term is 5n-3
n=1 5×1-3=2
n=2 5×2-3=7, etc
We write n terms both forward and backward as
Note that the next to last term 5(n-1)-3=5n-8
S=2 + 7 + 12 + ....

Let r= # of bowls of rice
b=# of bowls of broth
f=# of bowls of fowl
2r=3b=4f=# of guests
r+b+f=65
In terms of r we have b=(2/3)r, f=(1/2)r and so
r+(2/3)r+(1/2)r=65 solving for r we get
r=30 and so # of guests = 60

a is a first quadrant angle. sin(a)=8/17, the numbers are two sides of an 8,15,17 right triangle. cos(a)=15/17
b is a second quadrant angle. tan(b)=-7/24, the numbers are two sides of a 7,24,25 right triangle. The -7 is along the x-axis, the 24 is parallel to the y-axis. sin(b)=24/25,...

(sin2∅-1)/(tan2∅) = (1/sin2∅)-1
left side= -(1-sin2∅)/tan2∅
left side = -cos2∅/tan2∅
right side = (1/sin2∅)-1=(1-sin2∅)/sin2∅
right side = cos2∅/sin2∅=1/tan2∅
This is NOT an identity. Perhaps it needs revision.

w(4w-99)=13
4w²-99w-13=0
w=(99±√(99²-4×4×-13))/8
w=(99±√(10009))/8=(99±100.045)/8=24.8806 approximately
4w-99=h=.52249

3|m=11²-5k Let m=3n
11²-3n=5k
121-3×2=115=5×23 m=6, k=23
121-3×7=100=5×20 m=21, k=20
121-3×12=85=5×17 m=36, k=17
121-3×17=70=5×14 m=51, k=14
121-3×22=55=5×11 m=66, k=11
121-3×27=40=5×8 ...

The altitude of the trapezoid is 8 in, the two bases are 10 in and 18 in. The area is 1/2(18+10)×8=14×8=112 in²

|AUB|=50, |A-B|=20, |B-A|=15
AUB is the disjoint union of A-B, A∩B, and B-A
50 = 20+15+|A∩B| so |A∩B|=15
A=(A-B)∪(A∩B) a disjoint union |A|=20+15=35
B=(B-A)∪(A∩B) a disjoint union |B|=15+15=30

s=side of smaller square , s+3 = side of larger square
s2+(s+3)2=185
s2+s2+6s+9=185
2s2+6s-176=0
s2+3s-88=0
(s+11)(s-8)=0
s=-11 not possible or s=8
82+(8+3)2=64+121=185

Assume the table is a rectangle.
Let l = its length, w= its width
lw=39
2l=3w-5 so w=(2l+5)/3
l×(2l+5)/3=39
2l²+5l-117=0
(2l-13)(l+9)=0
The only positive root is l= 13/2=6 1/2 for the length
which makes the width w=6

This area is a triangle with the base along the y-axis from y=0 to y=4.
The altitude is the distance of the intersection of the lines y=x and y=4-3x, the point (1,1), from the y-axis
namely 1. The area is 4×1/2=2.
If you absolutely need to use calculus
Area is ∫01(4-4x)dx=...

I assume that what you meant was d=s+s2/20, which is
s2/20+s-d=0 or s2+20s-1500=0
(s-30)(s+50)=0 so that s=30, the negative result is rejected.

A+B+C=180
B=2A
C=A+20 substituting the values for B and C in the first equation we have
A+2A+A+20=180
4A+20=180 or 4A=160 gives A=40
B=2A=80
C=A+20=60

2N+5≥45 so that 2N≥40 or N≥20

The pattern you seem to be looking at is for every integer n, the number g you seek is such that
n/g = n+g
This gives g2+ng-n=0 or that g=(-n±√(n2+4n))/2
If you use the plus sign you get
n= 1 g=0.618033989
n=2 ...

After 1 year the car is worth 21000-.0937×21000=21000×.9063
After 2 years the car is worth (21000×.9063)×.9063
carry on