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3 years ago

12÷478 find two estimates plz help thank you ill give you 12 points

Mathematics

1 answer:

BaLLatris [955]3 years ago

6 0

10/480 equals about .02083 and 12/478 equals about <span>.02510. </span>

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Please help im so confused at math pls help

exis [7]

Answer:

where is the question ??

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2 years ago

Directions: Find the indicated trigonometric ratio as a fraction in simplest form. 1. Sin L =

Rainbow [258]

Answer:

$\sin L = 0.60$

$tan\ N = 1.33$

$\cos L = 0.80$

$\sin N = 0.80$

Step-by-step explanation:

Given

See attachment

From the attachment, we have:

$MN = 6$

$LN = 10$

First, we need to calculate length LM,

Using Pythagoras theorem:

$LN^2 = MN^2 + LM^2$

$10^2 = 6^2 + LM^2$

$100 = 36 + LM^2$

Collect Like Terms

$LM^2 = 100 - 36$

$LM^2 = 64$

$LM = 8$

Solving (a): $\sin L$

$\sin L = \frac{Opposite}{Hypotenuse}$

$\sin L = \frac{MN}{LN}$

Substitute values for MN and LN

$\sin L = \frac{6}{10}$

$\sin L = 0.60$

Solving (b): $tan\ N$

$tan\ N = \frac{Opposite}{Adjacent}$

$tan\ N = \frac{LM}{MN}$

Substitute values for LM and MN

$tan\ N = \frac{8}{6}$

$tan\ N = 1.33$

Solving (c): $\cos L$

$\cos L = \frac{Adjacent}{Hypotenuse}$

$\cos L = \frac{LM}{LN}$

Substitute values for LN and LM

$\cos L = \frac{8}{10}$

$\cos L = 0.80$

Solving (d): $\sin N$

$\sin N = \frac{Opposite}{Hypotenuse}$

$\sin N = \frac{LM}{LN}$

Substitute values for LM and LN

$\sin N = \frac{8}{10}$

$\sin N = 0.80$

7 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

Plzzzzzzzzzzzzzzzzzzz

inysia [295]

Answer:

Step-by-step explanation:

$\frac{12x+4}{4}\\ \\ 3x+1\\ \\ 3(3)+1\\ \\ 9+1\\ \\ 10$

7 0

2 years ago

Solve y = log3243. anything helps!

9966 [12]

Y=3.51
You could just plug it into your calculator!

3 0

2 years ago