Ask question

Ask question

All categories

3 years ago

9

Estimate the difference by rounding off to nearest tens . 6723 - 751

1 answer:

podryga [215]3 years ago

6 0

The answer is 5970 if I'm right

You might be interested in

Please help serious answers only

Stolb23 [73]

Answer: Maybe it's the third one I'm not completely sure. don't take my word for it.

Step-by-step explanation:

6 0

3 years ago

Bob works at a construction company. He has an equally likely chance to be assigned to work diffrent crews every day. He can be

Sergio [31]

The answer is 6 times

7 0

4 years ago

For the following telescoping series, find a formula for the nth term of the sequence of partial sums

gtnhenbr [62]

I'm guessing the sum is supposed to be

$\displaystyle\sum_{k=1}^\infty\frac{10}{(5k-1)(5k+4)}$

Split the summand into partial fractions:

$\dfrac1{(5k-1)(5k+4)}=\dfrac a{5k-1}+\dfrac b{5k+4}$

$1=a(5k+4)+b(5k-1)$

If $k=-\frac45$ , then

$1=b(-4-1)\implies b=-\frac15$

If $k=\frac15$ , then

$1=a(1+4)\implies a=\frac15$

This means

$\dfrac{10}{(5k-1)(5k+4)}=\dfrac2{5k-1}-\dfrac2{5k+4}$

Consider the $n$ th partial sum of the series:

$S_n=2\left(\dfrac14-\dfrac19\right)+2\left(\dfrac19-\dfrac1{14}\right)+2\left(\dfrac1{14}-\dfrac1{19}\right)+\cdots+2\left(\dfrac1{5n-1}-\dfrac1{5n+4}\right)$

The sum telescopes so that

$S_n=\dfrac2{14}-\dfrac2{5n+4}$

and as $n\to\infty$ , the second term vanishes and leaves us with

$\displaystyle\sum_{k=1}^\infty\frac{10}{(5k-1)(5k+4)}=\lim_{n\to\infty}S_n=\frac17$

7 0

3 years ago

Simplify each expression, #1: 4(f+2g) - 4g +f. #2: 5g(9m-2) + 3g

Serjik [45]

Answer:

1) 5f

2) 45gm-7g

Step-by-step explanation:

hope this helped!

7 0

3 years ago

What is the rate of change of the linear relationship modeled in the table (-2,5) (-1,4) (0,3) (1,2)

poizon [28]

The rate of change of the linear relationship is -1.

Explanation:

It is given that there is a linear relationship between all these points, these points lie on a straight line.

The equation to find the slope passing through two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Substituting the points $(-2,5)$ and $(-1,4)$ , we get,

$\begin{aligned}m &=\frac{4-5}{-1+2} \\&=\frac{-1}{1} \\&=-1\end{aligned}$

Thus, the slope is -1.

Thus, the rate of change of the linear relationship is -1.

8 0

4 years ago