Can someone help me with this? Thanks.
2 answers:
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2(4 + 2x) ≥ 5x + 5
First, we will need to expand our problem. Expanding is basically removing the parentheses. To do this, we will look at the first part of the problem to begin with. 2(4 + 2x). Since parentheses usually mean multiplication, we can start with 2(4). So, 2 × 4 = 8. We'll do the same thing with 2(2), 2 × 2 = 4.
Second, our next step is to subtract 4 from each side. We are trying to get the variable (x) on one side of the problem by itself.
Third, we can now simplify (5x) + 5 - (4). I put parentheses around what we are going to focus on. Subtract 5x - 4 to get 1, which can be put as the variable (x). Now we have, x + 5.
Fourth, let's subtract 5 from each side now. This will set up 8 - 5 which equals 3.
Fifth, we can switch sides now to get the result of this problem.
Answer:
First, we will need to expand our problem. Expanding is basically removing the parentheses. To do this, we will look at the first part of the problem to begin with. 2(4 + 2x). Since parentheses usually mean multiplication, we can start with 2(4). So, 2 × 4 = 8. We'll do the same thing with 2(2), 2 × 2 = 4.
Second, our next step is to subtract 4 from each side. We are trying to get the variable (x) on one side of the problem by itself.
Third, we can now simplify (5x) + 5 - (4). I put parentheses around what we are going to focus on. Subtract 5x - 4 to get 1, which can be put as the variable (x). Now we have, x + 5.
Fourth, let's subtract 5 from each side now. This will set up 8 - 5 which equals 3.
Fifth, we can switch sides now to get the result of this problem.
Answer:
Answer:
$5903.39
Step-by-step explanation:
This can be solve using compound interest formula. The formula is:
Where
F is the future amount (what we are looking for)
P is the present amount (which is 4800)
r is the rate of compound interest per year, in decimal (3% per year, 3/100 = 0.03)
t is the time in years ( t = 7)
Now we substitute these values into the formula and find F:
So, Colin would have $5903.39 after 7 years, in his account.