800 + 16x = 720 + 21x. Cancel out the 720 to get 80 + 16x = 21x. Subtract 16x from both sides to get 80 = 5x. Divide by 5 to get 16 = x, or x = 16. 16 hours. The charge is 800 + 16(16), = 720 + 21(16) which = 800 + 256 = 720 + 336 which = 1,056 = 1,056. Hope this helped!

6 0

3 years ago

A suit is sold for $68 while marked at $80. What<br> is the rate of discount?

MaRussiya [10]

15% is the rate of discount

85% of 80 is 68

And 100-15= 85.

6 0

2 years ago

Does anyone know how many ways you can make 16 and 19 on a grid

sasho [114]

You can look for the number 16 on the x-axis. (bottom line) and put a finger there. Then, look for 19 on the y -axis (left line) and put a finger there. Trace your left finger to the right and your right hand up until they meet. That is (16,19)

8 0

3 years ago

Prove or disprove (from i=0 to n) sum([2i]^4) <= (4n)^4. If true use induction, else give the smallest value of n that it doe

ddd [48]

Answer:

The statement is true for every n between 0 and 77 and it is false for $n\geq 78$

Step-by-step explanation:

First, observe that, for n=0 and n=1 the statement is true:

For n=0: $\sum^{n}_{i=0} (2i)^4=0 \leq 0=(4n)^4$

For n=1: $\sum^{n}_{i=0} (2i)^4=16 \leq 256=(4n)^4$

From this point we will assume that $n\geq 2$

As we can see, $\sum^{n}_{i=0} (2i)^4=\sum^{n}_{i=0} 16i^4=16\sum^{n}_{i=0} i^4$ and $(4n)^4=256n^4$ . Then,

$\sum^{n}_{i=0} (2i)^4 \leq(4n)^4 \iff \sum^{n}_{i=0} i^4 \leq 16n^4$

Now, we will use the formula for the sum of the first 4th powers:

$\sum^{n}_{i=0} i^4=\frac{n^5}{5} +\frac{n^4}{2} +\frac{n^3}{3}-\frac{n}{30}=\frac{6n^5+15n^4+10n^3-n}{30}$

Therefore:

$\sum^{n}_{i=0} i^4 \leq 16n^4 \iff \frac{6n^5+15n^4+10n^3-n}{30} \leq 16n^4 \\\\ \iff 6n^5+10n^3-n \leq 465n^4 \iff 465n^4-6n^5-10n^3+n\geq 0$

and, because $n \geq 0$ ,

$465n^4-6n^5-10n^3+n\geq 0 \iff n(465n^3-6n^4-10n^2+1)\geq 0 \\\iff 465n^3-6n^4-10n^2+1\geq 0 \iff 465n^3-6n^4-10n^2\geq -1\\\iff n^2(465n-6n^2-10)\geq -1$