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3 years ago

What is the approximate volume of the cone? use 3.14 for

Mathematics

1 answer:

Natali5045456 [20]3 years ago

8 0

Answer:

157 cm²

Step-by-step explanation:

volume = (1/3) * π * r² * h

volume = (1/3) * π * 5² * 6

volume = (1/3) * π * 25 * 6

volume = (1/3) * π * 150

volume = (1/3) * 471

volume = 157 cm²

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I NEED HELP PLEASE !!!!!

erastovalidia [21]

Answer:

c= (f-d)/ab

Step-by-step explanation:

abc+d = f
Subtract f from both sides so abc= f-d
Divide by ab from both sides so c = (f-d)/ab

8 0

3 years ago

Please answer this in steps....

SOVA2 [1]

<h2>Answer:</h2><h2>The answer is 3.14..</h2><h2>For finding it you have to divide the given numbers.</h2>

This is also a value of π (pi)

4 0

3 years ago

Traffic lights at three different junctions show green light at intervals of 10 seconds, 12 seconds and 15 seconds. They all sho

Flauer [41]

Answer:

it will show afain at 19 seconds

6 0

3 years ago

What is the equation of this line?<br><br>A: y=-3/2x<br>B: y=2/3x<br>C: y=-2/3x <br>D: y=3/2x

creativ13 [48]

Answer:

b) y = 2/3x

Step-by-step explanation:

Slope (m) formula: $\frac{y_2-y_1}{x_2-x_1}$

I will be using (3,2) and (-3,-2) to find the slope:

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

$y = mx + b\\y = \frac{2}{3}x+b$

(3,2)

$y = \frac{2}{3}x+b\\\\2 = \frac{2}{3}(3)+b\\\\2 = 2+b\\-2 -2\\\\0=b$

$y=mx+b\\\\y = \frac{2}{3}x+0\\\\y = \frac{2}{3}x$

Hope this helps!

8 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 $n \geq 0$ ,

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

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