All categories

3 years ago

What is the circumference of this circle given a radius of 7 in.?

Mathematics

2 answers:

EastWind [94]3 years ago

7 0

Hope this will help u...

Sedbober [7]3 years ago

4 0

Circumference is 43.98

You might be interested in

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

4 years ago

Which of the following are solutions to the system graphed below.

Kisachek [45]

Weird. I think you just need to look if the point falls on the shaded area. But only (-5,5) does ...

4 0

3 years ago

Determine the volume of the parallelepiped with one vertex at the origin and the three vertices adjacent to it at (2, â1, â1), (

valentinak56 [21]

Answer:

Step-by-step explanation:

Here is the complete question

Find the volume of the parallelepiped with one vertex at the origin and adjacent vertices at (1, 0, -3), (1, 2, 4), and (5, 1, 0).

Solution

We find the volume of the parallelepiped by making a 3 × 3 column matrix whose columns are the corresponding coordinates of the vertices of the parallelepiped.

So, (1, 0, -3), (1, 2, 4) and (5, 1, 0)

$A = \left[\begin{array}{ccc}1&1&5\\0&2&1\\-3&4&0\end{array}\right]$

The determinant of A is the volume of the parallelepiped. So,

detA = 1(2 × 0 - 4 × 1) - 1(0 × 0 - (-3) × 1) + 5(0 × 4 - (-3) × 2)

= 1(0 - 4) - 1(0 + 3) + 5(0 + 6)

= 1(-4) - 1(3) + 5(6)

= -4 - 3 + 30

= 23

So the volume of the parallelepiped is 23

4 0

3 years ago

Pam budgets $164 for fitness training. She buys a pair of hand weights for $12.65 and pays $10 per fitness class. Which inequali

NARA [144]

Answer:

Step-by-step explanation:

7 0

3 years ago

AO ⊥ (BCD)

Inga [223]

Answer:

V = 272.3 cube units

The surface area is 418.34 square units

4 0

3 years ago