What is the length of the diagonal?<br> 18 cm<br> 9 cm

Francisco is making an effort to drink more water, and has set a goal of 2 liters per day. He wants to know the volume of the wa

GarryVolchara [31]

Answer:

The volume of water bottle Francisco takes to school = 1.0048 liters

Step-by-step explanation:

The dimensions of the water bottle are given :

The Height of the water bottle is 20 cm

⇒ h = 20 cm

The Radius of the water bottle is 4 cm

⇒ r = 4 cm

⇒ Volume of the water bottle = π·r²·h

⇒ Volume = 3.14 × 4² × 20

⇒ Volume = 1004.8 cm³

Now, 1000 cm³ = 1 liter

⇒ 1004.8 cm³ = 1.0048 liter

Hence, The volume of water bottle Francisco takes to school = 1.0048 liters

3 0

3 years ago

Read 2 more answers

At the last minute, the technology company decides to expand the warehouse and asks the general contractor to double the dimensi

Lesechka [4]

<span>The expansion of the warehouse will not affect the dimensions of the surrounding fence.</span>

3 0

2 years ago

Can anyone help me solve all of this? Please help

disa [49]

They are all pretty easy all you do is multiply the cost by the amount of things you buy. For example number 4 your variables are cost (c) and songs (s). Your equation is c•s so $0.99 • s. 50 songs would be .99 • 50 which is $49.50

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 $n \geq 0$ ,

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