Answer:
B . . . . . .. . ....... .......
Answer:
167.36 weeks
Step-by-step explanation:
Given: Currently has $180, the cost is $860, and he plans to save $68 a week
To find: How many weeks will it take to save his money
Solution: If Barry currently has $180 and needs $860 for his tires also saves $68 a week, divide the amount he wants to save by the amount he saves a week, to find how many weeks will it take to save his money
860 ÷ 68 = 12.64
Subtract 180 because he already has $180 saved currently
180 - 12.64 = 167.36
Hence, it will take approximately 167 weeks to save his money up to $860
Answer:
6383.08
Step-by-step explanation:
J'ai dû utiliser google translate, mais j'espère que cela vous aidera.
Perhaps you know that

and

Then the problem is trivial, since

Then

so that <em>a</em> = 3 and <em>b</em> = 1.
Welcome to Brainly!
Binomial Theorem will follow this pattern:
The powers start at 3 and decrease down to 0 for the first term,
and start at 0 and increase to 3 for the other term.
(2x+4)^3 =

See how the power counts down on the (2x)
and counts up on the 4?
That's the pattern that our expansion must follow.
I left a little space in front of each term.
The coefficient in front of each term will come from the fourth row of Pascal's Triangle. The one that looks like this:
1 3 3 1
Those are the coefficients we want:

We can clean this up a little bit by getting rid of some of the junk. Anything to the 0th power is 1. So let's suppress all of our 1's because multiplying by 1 is not important.

Now apply exponent rule, distributing the power to both the 2 and the x where applicable.

Remember, multiplication is COMMUTATIVE, meaning we can multiply things in any order. So let's bring the numerical portion to the front of each term and multiply it all out.
