The answer, at least what I got, is 17.375 . They'd have to open an eighteenth package because the extra .375 is not in the original package, so I would go with B .
The answer is 15.
One-sixteenths is 1/16. Let's rewrite 15/16:
15/16 = (15 * 1) / 16 = 15 * 1/16
So, it is 15 times of one-sixteenths, or in other words 15 <span>sixteenths.</span>
Answer:
Step-by-step explanation:
Given is a system of equations as
We have 5 variables and 3 equations
a) coefficient matrix of this system is
1 -4 0 -1 0\\
0 1 0 -2 0\\
0 0 0 1 2\\
We find that x3 has no coefficient in any of the equations so we can omit x3 and write as equations for 4 variables as
1 -4 -1 0\\
0 1 -2 0\\
0 0 1 2\\
b) Augmented matrix is
1 -4 -1 0\\ 7
0 1 -2 0\\
3
0 0 1 2\\3
c) For row operations to ehelon form
we can do R1+4R2 = R1
We get
1 0 -9 0 \\ 19
0 1 -2 0 \\ 3
0 0 1 2 \\ 3
Now let us do R1 = R1+9R3 and R2 = R2+2R3
1 0 0 0 \\ 46
0 1 0 0 \\ 9
0 0 1 2 \\ 3
d) We find that there are infinite solutions to the system in parametric form, since x4 and x5 are linked with only one equation
e) x1 = 46, x2 = 9, x4+2x5 =3
Or x1 =46, x2 =9, x4 = 3-2x5, x5 = x5 is the parametric solution
Answer:
3xy^4+y-2/x
Step-by-step explanation:
12x^3y^4 + 4x^2y -8x
-----------------------------------
4x^2
We can break this fraction into pieces
12x^3y^4 4x^2y 8x
-------------- + --------- - ------------
4x^2 4x^2 4x^2
Taking the first piece
12/4 =3
x^3/x^2 =x
y^4/1 =y^4
3xy^4
Taking the second fractions
4/4=1
x^2/x^2 =1
y=y
y
Taking the third fraction
8/4=2
x/x^2 = 1/x
2/x
Adding them back together
3xy^4+y-2/x
Answer:
55mph
Step-by-step explanation:
car A distance, d1= 220
car b distance, d2= 260
speed of car A, s1= speed of car B- 10
let speed of car B be s2
s1= s2-10
speed = distance/time
time= distance/speed
As time of car A= time of car B
distance of Car A/speed of Car A=distance of Car B/speed of Car B
220/s2-10= 260/s2
s2=(s2-10)260/220
s2=65
s1= s2- 10
= 65-10
=55
hence speed of first car is 55mph !
