First you have to double the simple interest to get the amount earned in 1 year. That gives $3.56 earned and a total balance of $893.56
r = (1/t)(A/P - 1) t=time, A=amount of money with interest, P=amount invested
r = (1/1)((893.56/890) - 1) = 0.004
r = 0.004
now convert to a decimal to get .04 = 4%
Answer:
$31.5
Step-by-step explanation:
-$45
-30 percent off of $45
30 times 45 divided by 100
1350/100
13.5
45-13.5
$31.5
Answer:
i- umm- its, well i got two answers n this question as well, but umm, i think its
1 5/12 maybe
Step-by-step explanation:
Answer:
I cant see anything on that picture. Wish I could help bruv. I wish I could help. It's a shame innit.
Answer:
Exponential decay.
Step-by-step explanation:
You can use a graphing utility to check this pretty quickly, but you can also look at the equation and get the answer. Since the function has a variable in the exponent, it definitely won't be a linear equation. Quadratic equations are ones of the form ax^2 + bx + c, and your function doesn't look like that, so already you've ruled out two answers.
From the start, since we have a variable in the exponent, we can recognize that it's exponential. Figuring out growth or decay is a little more complicated. Having a negative sign out front can flip the graph; having a negative sign in the exponent flips the graph, too. In your case, you have no negatives; just 2(1/2)^x. What you need to note here, and you could use a few test points to check, is that as x gets bigger, (1/2) will get smaller and smaller. Think about it. When x = 0, 2(1/2)^0 simplifies to just 2. When x = 1, 2(1/2)^1 simplifies to 1. Already, we can tell that this graph is declining, but if you want to make sure, try a really big value for x, like 100. 2(1/2)^100 is a value very very very veeery close to 0. Therefore, you can tell that as the exponent gets larger, the value of the function goes down and gets closer and closer to zero. This means that it can't be exponential growth. In the case of exponential growth, as the exponent gets bigger, your output should increase, too.
