If a = first term and r = common ratio we have
a + ar + ar^2 = 13 and ar^2 / a = r^2 = 9
so r = 3
and a + 3a + 9a = 13
so a = 1
so they are 1,3 and 9
2.
in geometric series we have
4 , 4r ,4r^2 , 60
Arithmetic;
4, 4r , 4r + d , 4r + 2d
so we have the system of equations
4r + 2d = 60
4r^2 = 4r + d
From first equation
2r + d = 30
so d = 30 - 2r
Substitute for d in second equation:-
4r^2 - 4r - (30-2r) = 0
4r^2 - 2r - 30 =0
2r^2 - r - 15 = 0
(r - 3)(2r + 5) = 0
r = 3 or -2.5
r must be positive so its = 3
and d = 30 - 2(3) = 24
and the numbers are 4*3 = 12 , 4*3^2 = 36
first 3 are 4 , 12 and 36 ( in geometric)
and last 3 are 12, 36 and 60 ( in arithmetic)
The 2 numbers we ause are 12 and 36.
do you need to include the wiggle infront of the first bracket? if not;
(x+1) ÷ [(x^2+2) x (2x-3dx)]
x^2 x 2x = 2x^3
x^2 x -3dx = -3dx^3
2 x 2x = 4x
2 x -3dx = - 6dx
i cant find a way to make it equal 0 so i think the answer is just
x+1 over 2x^3 - 3dx^3 + 4x - 6dx as a fraction
Surface area = 2(ab+bc+ac)
a=5.2 ft, b=2.4 ft, c=3.5ft
Surface area = 2(5.2 * 2.4 + 2.4 * 3.5 + 3.5 * 5.2) =
2(12.48 + 8.4 + 18.2) =
2 * 39.08 =
78.16 ≈ 78.2 ft² ← <span>to the nearest tenth</span>
Answer:
Increase
Step-by-step explanation:
The researcher records the following estimates: 450, 426, 310, 500, and 220.
The mean of these estimates is derived below.
Mean = (450+426+310+500+220)/5
=1906/5=381.2
If the researcher removes the estimate of 220.
The mean of the other numbers will be:
Mean =(450+426+310+500)/4
=1686/4=421.5
By comparison of the two mean, we can see that the value of the mean will increase.
Answer: 1 1/2
Step-by-step explanation:
First subtract 1 1/3 from 3 1/3 to find what is left over from her first batch of browns, 2. Then subtract 1 1/3 from 2 and get 2/3. 2/3 is 1/2 of 1 1/3, so she can make 1 1/2 more batches of brownies.
