A triangle has sides with lengths 17, 12, and 9. Is the triangle right, acute, or obtuse?

Ow dude my freaking neck hurts

Xelga [282]

Answer:

not funny

didnt laugh

Step-by-step explanation:

6 0

3 years ago

M + w = 27<br> m+ 7 = w<br> I need help on how to do this plzz help ASAP

snow_tiger [21]

M=20 since 27-m=7 it means that 20+7=27

4 0

3 years ago

Chelsea completes 35 math problems in 7 min.

Andru [333]

Answer = 5 problems per minute

Because you do the number of problems divided by the minutes eg 35/7 = 5

3 0

4 years ago

The position of a large fish relative to the water surface was −17 1/3 feet. It suddenly dove straight down and increased its de

boyakko [2]

So let’s imagine this on a number line…so we know that the fish starts at -17 1/3 on the number line but he dives (that means go lower in our number line) 6 1/2 feet downward so let’s set up an equation:

-17 1/3 - 6 1/2
Now, let’s subtract the two integers by finding the least common denominator of 2 and 3 so it can be easier to solve in this case it’s 6…then let’s set up our new equation:

-17 2/6 - 6 3/6
We know it’s a negative number since we are subtracting a negative number and our final answer is -23 5/6

6 0

2 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 $n \geq 0$ ,

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