-<br> At x=3 the function given by f(x) {x^2, x>3<br> 6x-9x>3<br> =

1. In 2011, Molly and Jerry both went to

True [87]

Answer:

2017

Step-by-step explanation:

Molly goes every 2 years

Jerry goes every 3 years

LCM(2,3) = 2 x 3 = 6

2011 + 6 = 2017

The answer is 2017

6 0

2 years ago

This is the question

AnnZ [28]

Answer:

cool question I love it

3 0

3 years ago

Read 2 more answers

Find the second derivative y= x^2lnx

Dimas [21]

First we use product rule
y=x^2lnx
dy/dx = x^2 d/dx (lnx) + lnx d/dx (x^2)
dy/dx = x^2 (1/x) + lnx (2x)
dy/dx = x + 2xlnx

now taking second derivative:
d2y/dx2 = 1 + 2[x (1/x) + lnx (1)]
d2y/dx2 = 1 + 2[1+lnx]
1+2+2lnx
3+2lnx is the answer

3 0

3 years ago

Which expression is equivalent to *picture attached*

DiKsa [7]

Answer:

The correct option is;

$4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3 \left (\dfrac{50(51) }{2} \right )$

Step-by-step explanation:

The given expression is presented as follows;

$\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3 \right )$

Which can be expanded into the following form;

$\sum\limits _{n = 1}^{50} \left (4\cdot n^2 + 3 \cdot n\right ) = 4 \times \sum\limits _{n = 1}^{50} \left n^2 + 3 \times\sum\limits _{n = 1}^{50} n$

From which we have;

$\sum\limits _{k = 1}^{n} \left k^2 = \dfrac{n \times (n+1) \times(2n+1)}{6}$

$\sum\limits _{k = 1}^{n} \left k = \dfrac{n \times (n+1) }{2}$

Therefore, substituting the value of n = 50 we have;

$\sum\limits _{n = 1}^{50} \left k^2 = \dfrac{50 \times (50+1) \times(2\cdot 50+1)}{6}$

$\sum\limits _{k = 1}^{50} \left k = \dfrac{50 \times (50+1) }{2}$

Which gives;

$4 \times \sum\limits _{n = 1}^{50} \left n^2 = 4 \times \dfrac{n \times (n+1) \times(2n+1)}{6} = 4 \times \dfrac{50 \times (50+1) \times(2 \times 50+1)}{6}$

$3 \times\sum\limits _{n = 1}^{50} n = 3 \times \dfrac{n \times (n+1) }{2} = 3 \times \dfrac{50 \times (51) }{2}$

$\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3 \right ) = 4 \times \dfrac{50 \times (50+1) \times(2\times 50+1)}{6} +3 \times \dfrac{50 \times (51) }{2}$

Therefore, we have;

$4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3 \left (\dfrac{50(51) }{2} \right )$ .

4 0

3 years ago

Explain how to find which value is greater (5/6)*81 or (9/10)*81 without multiplying?

8_murik_8 [283]

The value (9/10)*81 is greater than (5/6)*81

<h3>How to determine the greater value?</h3>

The values are given as:

(5/6)*81 and (9/10)*81

Remove the common factor in both values

So, we have:

5/6 and 9/10

Convert to decimals

0.83 and 0.9

0.9 is greater than 0.83

Hence, (9/10)*81 is greater than (5/6)*81

- At x=3 the function given by f(x) {x^2, x>3 6x-9x>3 =