MaxWong

I think heureka solved this problem in a detailed manner in this post here: 

If you need an answer to another problem, please start another post.

Expand out. 

\(\quad(x + 2)(3x^2 - x + 8)\\ = x(3x^2 - x + 8) + 2(3x^2 - x + 8)\\ = 3x^3 - x^2 + 8x + 6x^2 - 2x + 16\\ = 3x^3 + 5x^2 + 6x + 16\)

Now you can compare the coefficients of \(3x^3 + 5x^2 + 6x + 16\) and \(Ax^3 + Bx^2 + Cx + D\) to find A, B, C, D. Then just calculate A + 2B + 3C + 4D.

Let t minutes be the time needed.

Mike is traveling at \(\dfrac{71}{60}\text{ mi/min}\) and Ike is traveling at \(1\text{ mi/min}\).

1 minute after Ike passed the one-mile mark, Mike passes the one-mile mark. But in that one minute that Mike used to catch up to Ike, Ike has traveled an extra 1 mile. So at that specific moment:

|-------------------------|------------------------|

            1 mi            Mike        1 mi            Ike

t minutes after that moment, total distance travelled by Mike is \(\left(1 + \dfrac{71}{60}t\right)\) miles and total distance travelled by Ike is \((2 + t)\) miles.

When Mike catches up to Ike, the total distance traveled will become the same. so \(1 + \dfrac{71}{60}t = 2+ t\). Please solve for t on your own.

Then (t + 2) minutes is the answer.

Let x = 3^c. Then \(3x^2 = 28x\). You can use quadratic formula or factorization to solve for x. Can you take it from here?

Percentage increase = \(\dfrac{\text{new} - \text{old}}{\text{old}} \times 100\%\). Plug in new = 119 and old = 85 and you will get the answer.

https://web2.0calc.com/questions/i-need-help-fast_10#r1

Same question here

\((3x^3 - 3x + 4) - (x^3 - 2) = 3x^3 - 3x + 4 - x^3 + 2\)

Now you just collect the like terms and it's done. Remember to order your terms in descending order of power.

I will do (a) for you as an example.

x + floor(x) = 1

x = 1 - floor(x), which is an integer.

We conclude that x is an integer. Then floor(x) = x.

x + x = 1

2x = 1

x = 1/2 (rejected)

1/2 is rejected because it is not an integer.

There are no such x >= 0 such that f(x) = 1.

(b) and (c) are similar, and there are infinitely many possible values of f(x) for 0 <= x <= 10, if you consider all real numbers in that range.

10 * 11 * 12 * ... * 25 = (10 * 15 * 20 * 25) * (other numbers)

The other numbers don't have any factor of 5. We consider 10 * 15 * 20 * 25.

10 * 15 * 20 * 25 = 5^5 * 2^3 * 3

Pick two even numbers from (other numbers) to make 5^5 * 2^5 = 10^5.

There will be 5 zeros at the end.