Let x be the mass of the paperbacks and y be the mass of the textbook.
20x + 9y = 44.4 ----------- (1)
25x + 10y = 51 -------------(2)
(1) x 10:
200x + 90y = 444 --------(1a)
(2) x 9:
225x + 90y = 459 --------(2a)
(2a) - (1a):
25x = 15
x = 0.6 -------- sub into (1)
20 (0.6) + 9y = 44.4
12 + 9y = 44.4
9y = 44.4 - 12
9y = 32.4
y = 3.6
So the paperback's mass is 0.6 pounds and textbook is 3.6 pounds
Answer:
Step-by-step explanation:
A=3
B=4
C=6
D=5
E=1
F=7
G=2
Answer:
Step-by-step explanation:
We first have to write the equation for the sequence, then finding the first five terms will be easy. Follow the formatting:
and we are given enough info to fill in:
and
and
or in linear format:
where n is the position of the number in the sequence. We already know the first term is -35.
The second term:
so
and
The third term:
and
so
and we could go on like this forever, but the nice thing about this is when we know the difference all we have to do is add it to each number to get to the next number.
That means that the fourth term will be -27 + 4 which is -23.
The fifth term then will be -23 + 4 which is -19. You can check yourself by filling in a 5 for n in the equation and solving:
and
so
It must be an obtuse triangle
First term ,a=4 , common difference =4-7=-3, n =50
sum of first 50terms= (50/2)[2×4+(50-1)(-3)]
=25×[8+49]×-3
=25×57×-3
=25× -171
= -42925
derivation of the formula for the sum of n terms
Progression, S
S=a1+a2+a3+a4+...+an
S=a1+(a1+d)+(a1+2d)+(a1+3d)+...+[a1+(n−1)d] → Equation (1)
S=an+an−1+an−2+an−3+...+a1
S=an+(an−d)+(an−2d)+(an−3d)+...+[an−(n−1)d] → Equation (2)
Add Equations (1) and (2)
2S=(a1+an)+(a1+an)+(a1+an)+(a1+an)+...+(a1+an)
2S=n(a1+an)
S=n/2(a1+an)
Substitute an = a1 + (n - 1)d to the above equation, we have
S=n/2{a1+[a1+(n−1)d]}
S=n/2[2a1+(n−1)d]
