If a matrix is invertible, then its determinant is not zero.
For an invertible matrix A, the equation above reads, with B=A-1
det(AA-1)=detAdetA-1. Since AA-1=I, the identity matrix, whose determinant is 1, we have
1=detAdetA-1 or that detA-1=1/detA.
subtract twice the first equation from the second
subtract the first equation from the third
y=1 from third equation
Let x be quantity of 20% alcohol
Let y be quantity of 60% alcohol
The number of ways of choosing one person from each group is 5C1×8C1×5C1×10C1×4C1× 6C1=48000
The number of ways of choosing six people from 38 people is 38C6=2760681
The probability in question is the ratio 48000/2760681
Let x=1+.072/12=1.006 the monthly rate of increase of money
Let F be the future value of the cash stream.
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...
That's as far as I can go without seeing the choices
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
It is necessary to assume that x1 and x0 are known. This makes the fact that x2=x1-x0 given.
Let S be the sum in question. Each term is 5n-3
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
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.
3|m=11²-5k Let m=3n
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
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
s=-11 not possible or s=8
Assume the table is a rectangle.
Let l = its length, w= its width
2l=3w-5 so w=(2l+5)/3
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.