Ask question

Ask question

All categories

3 years ago

12

A child standing on a river bank ten feet above the river throws a toward the river at a speed of 12 feet per second . How long

does it take before the rock splashes into the river ? Use the equation h=-16t^ 2 +vt+s
PLEASESE HELP

1 answer:

Anastasy [175]3 years ago

8 0

sure just give me head

You might be interested in

What is 20/8 in the lowest mixed form?

Kisachek [45]

20/8 = 10/4 = 5/2

So, it's mixed form would be 2 1/2

8 0

3 years ago

Help please<br>urgent!!!!!!

snow_tiger [21]

The Chicken, in total, cost $4.56

You need to figure out the total in pounds, which is 4 3/4. Then multiply that by the price, .96

6 0

3 years ago

1) A boat traveled 448 miles downstream and back. The trip downstream took 14 hours. The trip

Serhud [2]

Answer:B

I think we are in the same class lol

Step-by-step explanation:448 divided by 14= 32

448 divided by 16= 28

B=speed of boat in water

Y= Speed of current

B+y=32

B-y=28

32 + 28 =60

60 divided by 2 = 30

32 -30 =2

B=30

Y=2

Add my discord if you need more help Snapple#0141

6 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 $n \geq 0$ ,

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

Observe that, because $n \geq 2$ and is an integer,

$n^2(465n-6n^2-10)\geq -1 \iff 465n-6n^2-10 \geq 0 \iff n(465-6n) \geq 10\\\iff 465-6n \geq 0 \iff n \leq \frac{465}{6}=\frac{155}{2}=77.5$

In concusion, the statement is true if and only if n is a non negative integer such that $n\leq 77$

So, 78 is the smallest value of n that does not satisfy the inequality.

Note: If you compute $(4n)^4- \sum^{n}_{i=0} (2i)^4$ for 77 and 78 you will obtain:

$(4n)^4- \sum^{n}_{i=0} (2i)^4=53810064$

$(4n)^4- \sum^{n}_{i=0} (2i)^4=-61754992$

7 0

3 years ago

What is the equation of the line passing through the points (3,6) and (2,10)

andrezito [222]

Answer: y= - 4x+18

Step-by-step explanation:

Equation: y=mx+b

***remember: b is the y-intercept and m is the slope.

m= $\frac{y2-y1}{x2-x1}$

3= x1

2= x2

6= y1

10=y2

m= $\frac{10-6}{2-3}$ = $\frac{4}{1}$ = -4

m=-4

Now we have y=-4x+b , so let's find b.

You can use either (x,y) such as (3,6) or (2,10) point you want..the answer will be the same:

(3,6). y=mx+b or 6=-4 × 3+b, or solving for b: b=6-(-4)(3). b=18.

(2,10). y=mx+b or 10=-4 × 2+b, or solving for b: b=10-(-4)(2). b=18.

Equation of the line: y=-4x+18

3 0

3 years ago