All categories

3 years ago

Can Wendell make a rectangular garden with the pieces of wood he has? If yes, which pieces can he use?

Mathematics

1 answer:

Mamont248 [21]3 years ago

5 0

Answer:nope

Step-by-step explanation:

You might be interested in

Please calculate this limit <br>please help me

Tasya [4]

Answer:

We want to find:

$\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n}$

Here we can use Stirling's approximation, which says that for large values of n, we get:

$n! = \sqrt{2*\pi*n} *(\frac{n}{e} )^n$

Because here we are taking the limit when n tends to infinity, we can use this approximation.

Then we get.

$\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n} = \lim_{n \to \infty} \frac{\sqrt[n]{\sqrt{2*\pi*n} *(\frac{n}{e} )^n} }{n} = \lim_{n \to \infty} \frac{n}{e*n} *\sqrt[2*n]{2*\pi*n}$

Now we can just simplify this, so we get:

$\lim_{n \to \infty} \frac{1}{e} *\sqrt[2*n]{2*\pi*n} \\$

And we can rewrite it as:

$\lim_{n \to \infty} \frac{1}{e} *(2*\pi*n)^{1/2n}$

The important part here is the exponent, as n tends to infinite, the exponent tends to zero.

Thus:

$\lim_{n \to \infty} \frac{1}{e} *(2*\pi*n)^{1/2n} = \frac{1}{e}*1 = \frac{1}{e}$

7 0

3 years ago

4 (5x-3) -64 (3-x) -3 (12x-4)=96

Yuki888 [10]

Are you trying to simply the equation if so x = 6

4 0

3 years ago

A juggler tosses a ball into the air . The balls height, h and time t seconds can be represented by the equation h(t)= -16t^2+40

malfutka [58]

PART A

The given equation is

$h(t) = - 16 {t}^{2} + 40t + 4$

In order to find the maximum height, we write the function in the vertex form.

We factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + 4$

We add and subtract

$- 16(- \frac{5}{4} )^{2}$

to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + - 16( - \frac{5}{4})^{2} - -16( - \frac{5}{4})^{2} + 4$

We again factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4})^{2} ) - -16( - \frac{5}{4})^{2} + 4$

This implies that,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4}) ^{2} ) + 16( \frac{25}{16}) + 4$

The quadratic trinomial above is a perfect square.

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +25+ 4$

This finally simplifies to,

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +29$

The vertex of this function is

$V( \frac{5}{4} ,29)$

The y-value of the vertex is the maximum value.

Therefore the maximum value is,

$29$

PART B

When the ball hits the ground,

$h(t) = 0$

This implies that,

$- 16 ( t- \frac{5}{4}) ^{2} +29 = 0$

We add -29 to both sides to get,

$- 16 ( t- \frac{5}{4}) ^{2} = - 29$

This implies that,

$( t- \frac{5}{4}) ^{2} = \frac{29}{16}$

$t- \frac{5}{4} = \pm \sqrt{ \frac{29}{16} }$

$t = \frac{5}{4} \pm \frac{ \sqrt{29} }{4}$

$t = \frac{ 5 + \sqrt{29} }{4} = 2.60$

or

$t = \frac{ 5 - \sqrt{29} }{4} = - 0.10$

Since time cannot be negative, we discard the negative value and pick,

$t = 2.60s$

8 0

3 years ago

2 and one half divide by one fourth

solmaris [256]

Hi,

$2 \frac{1}{2} \div \frac{1}{4} \\ 2 \frac{2}{4} \div \frac{1}{4} = 10$

Hope this helps.
r3t40

3 0

4 years ago

If the data analyst hypothesizes that grocery sales are lower on Sunday than on other days, the null hypothesis would likely sta

nlexa [21]

Answer:

The answer is "Grocery sales are lower or no different on Sunday than on other days"

Step-by-step explanation:

Its null hypothesis does not always consider any discrepancy between both the comparable qualities throughout facts and figures. Furthermore, researchers recommend also have lower or even no opposite transactions for Sunday than with other days, because there is no connection between grocery sales and Saturdays and Sundays.

7 0

3 years ago