All categories

3 years ago

Please Help!!!!!!!

Mathematics

2 answers:

Umnica [9.8K]3 years ago

7 0

Answer:

180?

Step-by-step explanation:

Likurg_2 [28]3 years ago

4 0

Answer: The new beehive will hold 200 pounds of honey

Explanation: 30 divided by 6 is five so 40*5 is 200

You might be interested in

Factor the polynomial function over the complex numbers. f(x)=x^3-2x^2+3x-6

Airida [17]

$f(x)=x^3-2x^2+3x-6=x^2(x-2)+3(x-2)=(x-2)(x^2+3)\\\\i=\sqrt{-1}\to i^2=-1\\\\x^2+3=x^2-(-3)=x^2-3i^2=x^2-(\sqrt3)^2i^2)=x^2-(i\sqrt3)^2\\\\\text{use}\ a^2-b^2=(a-b)(a+b)\\\\\boxed{f(x)=(x-2)(x-i\sqrt3)(x+i\sqrt3)}$

6 0

3 years ago

Hi ! How to solve this question?

tester [92]

You should actually have

$-x^2-4x-10=-(x^2+4x+10)=-(x^2+4x+4+6)=-((x+2)^2+6)=-(x+2)^2-6$

Now, remember that $x^2$ is always non-negative, so $(x+2)^2\ge0$ for any value of $x$ . This means $-(x+2)^2\le0$ for any $x$ , and so

$-(x+2)^2-6\le0-6=-6$

i.e. f(x) is at most -6, and hence negative for all $x$ .

4 0

3 years ago

Find the length of the shorter leg of a right triangle if the longer leg is 7 feet more than the shorter leg and the hypotenuse

gtnhenbr [62]

You can use the pythagorean theorem.
$a^{2} + b^{2} = c^{2}$
Now lets write out what we know.

Shorter leg = x
Longer Leg = x + 7
Hypotenuse = x - 7

$a^{2} + b^{2} = c^{2}$
Now solve for x
$x^{2} + (x+7)^{2} = (2x-7)^{2}$
x = 28
Since you know x = 28, which is the shorter leg, you can solve for the rest by just plugging in 28 where x is located. For instance, longer let is x + 7 so
28 + 7 = 35

6 0

3 years ago

Need help with number 1 and 2

Hatshy [7]

Me das mas información?

8 0

3 years ago

Let f(x) = [infinity] xn n2 n = 1. find the intervals of convergence for f. (enter your answers using interval notation. ) find

inna [77]

Best guess for the function is

$\displaystyle f(x) = \sum_{n=1}^\infty \frac{x^n}{n^2}$

By the ratio test, the series converges for

$\displaystyle \lim_{n\to\infty} \left|\frac{x^{n+1}}{(n+1)^2} \cdot \frac{n^2}{x^n}\right| = |x| \lim_{n\to\infty} \frac{n^2}{(n+1)^2} = |x| < 1$

When $x=1$ , $f(x)$ is a convergent $p$ -series.

When $x=-1$ , $f(x)$ is a convergent alternating series.

So, the interval of convergence for $f(x)$ is the closed interval $\boxed{-1 \le x \le 1}$ .

The derivative of $f$ is the series

$\displaystyle f'(x) = \sum_{n=1}^\infty \frac{nx^{n-1}}{n^2} = \frac1x \sum_{n=1}^\infty \frac{x^n}n$

which also converges for $|x|$ by the ratio test:

$\displaystyle \lim_{n\to\infty} \left|\frac{x^{n+1}}{n+1} \cdot \frac n{x^n}\right| = |x| \lim_{n\to\infty} \frac{n}{n+1} = |x| < 1$

When $x=1$ , $f'(x)$ becomes the divergent harmonic series.

When $x=-1$ , $f'(x)$ is a convergent alternating series.

The interval of convergence for $f'(x)$ is then the closed-open interval $\boxed{-1 \le x < 1}$ .

Differentiating $f$ once more gives the series

$\displaystyle f''(x) = \sum_{n=1}^\infty \frac{n(n-1)x^{n-2}}{n^2} = \frac1{x^2} \sum_{n=1}^\infty \frac{(n-1)x^n}{n} = \frac1{x^2} \left(\sum_{n=1}^\infty x^n - \sum_{n=1}^\infty \frac{x^n}n\right)$

The first series is geometric and converges for $|x|$ , endpoints not included.

The second series is $f'(x)$ , which we know converges for $-1\le x$ .

Putting these intervals together, we see that $f''(x)$ converges only on the open interval $\boxed{-1 < x < 1}$ .

6 0

2 years ago