Answer the following: 5x6+3%5-12+3

Mathematics

2 answers:

QveST [7]3 years ago

7 0

=5x ( exponent 6 beside x) + -117/20

Softa [21]3 years ago

4 0

Answer: 24

Step-by-step explanation:

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If y varies directly as x and y=3 when x=2 find y when x=8

satela [25.4K]

Answer:

Step-by-step explanation:

x=2 was multiplied by 4 to get to x=8 so you multiply y=3 by 4 to make it y=12

6 0

3 years ago

Read 2 more answers

1. Describe a situation that could be modeled by x≤50

Dimas [21]

There are many ways to answer. The x represents some number. We don't know what the number is, but we know that it must be less than or equal to 50. Put another way, the number can be anything you want as long as it doesn't go past 50. We say that 50 is the so called "ceiling" more or less.

Some examples:

* An elevator can only hold 50 people at maximum. Therefore, x can be any number smaller than 50 or 50 itself. Having 51 or over will be too much.

* You can only work 50 hours for one stretch of some 2 week period. If x is the number of hours you work, then x must be 50 or less as written by $x \le 50$ . So x could be x = 37 as it's less than 50, but x = 62 is not possible.

* For some small ride at a theme park, the seats are designed such that only people 50 inches or less can ride on them. If x is the height of a person in inches, then $x \le 50$ means something like x = 37 is possible but x = 62 is too high.

8 0

3 years ago

Use the Taylor series you just found for sinc(x) to find the Taylor series for f(x) = (integral from 0 to x) of sinc(t)dt based

Marina CMI [18]

In this question (brainly.com/question/12792658) I derived the Taylor series for $\mathrm{sinc}\,x$ about $x=0$ :

$\mathrm{sinc}\,x=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}$

Then the Taylor series for

$f(x)=\displaystyle\int_0^x\mathrm{sinc}\,t\,\mathrm dt$

is obtained by integrating the series above:

$f(x)=\displaystyle\int\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}\,\mathrm dx=C+\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}$

We have $f(0)=0$ , so $C=0$ and so

$f(x)=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}$

which converges by the ratio test if the following limit is less than 1:

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(-1)^{n+1}x^{2n+3}}{(2n+3)^2(2n+2)!}}{\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}}\right|=|x^2|\lim_{n\to\infty}\frac{(2n+1)^2(2n)!}{(2n+3)^2(2n+2)!}$