X= 3 y=3
(3)2x+y=9
4x-3y=3
6x+3y=27
4x-3y=3
——————
10x=30
—————-
10
x=3
2(3)+y=9
6+y=9
-6. -6
y=3
A, C, and D are all correct answers, since you are taking away 4 in each of the answers. Hope this helped!
Answer:
- as written, c ≈ 0.000979 or c = 4
- alternate interpretation: c = 0
Step-by-step explanation:
<em>As written</em>, you have an equation that cannot be solved algebraically.
(32^2)c = 8^c
1024c = 8^c
1024c -8^c = 0 . . . . . . rewrite as an expression compared to zero
A graphical solution shows two values for c: {0.000978551672551, 4}. We presume you're interested in c = 4.
___
If you mean ...
32^(2c) = 8^c
(2^5)^(2c) = (2^3)^c . . . . rewriting as powers of 2
2^(10c) = 2^(3c) . . . . . . . simplify
10c = 3c . . . . . . . . . . . . . .log base 2
7c = 0 . . . . . . . . . . . . . . . subtract 3c
c = 0 . . . . . . . . . . . . . . . . divide by 7
Answer:
Use PEMDAS so division would be the first step
Step-by-step explanation:
1)Parentheses
2)Exponents
3)Multiplication
4)Division
5)Addition
6)Subtraction
<span>Let a_0 = 100, the first payment. Every subsequent payment is the prior payment, times 1.1. In order to represent that, let a_n be the term in question. The term before it is a_n-1. So a_n = 1.1 * a_n-1. This means that a_19 = 1.1*a_18, a_18 = 1.1*a_17, etc. To find the sum of your first 20 payments, this sum is equal to a_0+a_1+a_2+...+a_19. a_1 = 1.1*a_0, so a_2 = 1.1*(1.1*a_0) = (1.1)^2 * a_0, a_3 = 1.1*a_2 = (1.1)^3*a_3, and so on. So the sum can be reduced to S = a_0 * (1+ 1.1 + 1.1^2 + 1.1^3 + ... + 1.1^19) which is approximately $5727.50</span>
