<span>Form a sequence that has two arithmetic means between -13 and 89. a. -13, 33, 43, 89 c. -13, 21, 55, 89 b. -18, -36, -72, -144 d. -18, -81, -144
Solution:
Since it has to be between -13 and 89, letter d and b are not anymore considered to be the answer.
for a:
33-(-13)=46=d
43-33=10=d the value for this d is different from the two sequence,
89-43=46=d
they have different value for d, thus this is not the answer!
for c:
21-(-13)=34=d
55-21=34=d
89-55=34=d
they have the same value for d, thus the correct answer is </span><span> c. -13, 21, 55, 89</span>
(193/400 )x100 equals 48.25 %
a(x+1)(x-1)+b(x-2)(x+1)+(x+1)^2=9x^2-x-10
(x+1)(a(x-1)+b(x-2)+(x+1))=(x+1)(9x-10)
a(x-1)+b(x-2)+x+1=9x-10
Now this equation is much simpler!
(a+b)x-a-2b+x+1=9x-10
(a+b)x-a-2b=8x-11
(a+b-8)x-a-2b-11=0
a+b-8=(a+2b-11)/x
I can't solve it 3 variables and 1 equations means infinite answers so yea.
Answer:
21z+84
Step-by-step explanation.
7x3=21 then put the z at the end 21z.
7x12=84. Put it after the 21z. 21z+84
Answer:
slope i think
Step-by-step explanation:
