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3 years ago

5

Round to the thousandth place: 82.6925

1 answer:

Vinvika [58]3 years ago

4 0

82.693

the rule is when its 4 and under round down and 5 and up round up

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The expression can be simplified and written without negative exponents as...

Ksju [112]

$\dfrac{(27y^{-2})^\frac{1}{3}}{y^{-\frac{2}{3}}}=(27)^\frac{1}{3}(y^{-2})^\frac{1}{3}:\dfrac{1}{y^\frac{2}{3}}=\sqrt[3]{27}y^{-2\cdot\frac{1}{3}}\cdot y^\frac{2}{3}\\\\=3y^{-\frac{2}{3}}\cdot y^\frac{2}{3}=3y^{-\frac{2}{3}+\frac{2}{3}}=3y^0=3\\\\\text{used:}\\\\(a\cdot b)^n=a^n\cdot b^n\\\\a^\frac{1}{n}=\sqrt[n]{a}\\\\a^{-n}=\dfrac{1}{a^n}\\\\(a^n)^m=a^{n\cdot m}\\\\a^n\cdot a^m=a^{n+m}\\\\a^0=1,\ a\neq0$

6 0

4 years ago

Help please?!?!?!?!?!?!

svp [43]

slope is change in y over change in x

use 2 points from the table so -3,-21 and -6,-39

change in Y: -39 - -21 = -18

change in x = -6 - -3 = -3

slope = -18/-3 = 6

slope = 6

8 0

3 years ago

A particle moves along a line with acceleration a (t) = -1/(t+2)2 ft/sec2. Find the distance traveled by the particle during the

amm1812

Answer:

$s(t)=(ln|7|+ln|2|)\,ft$

Step-by-step explanation:

Acceleration is second derivative of distance and are related as:

$a(t)=\frac{d^2s}{dt^2}\\\\\frac{d^2s}{dt^2}=\frac{-1}{(t+2)^2}\\$

Integrating both sides w.r.to t

$v(t)=\frac{ds}{dt}=\frac{1}{t+2} +C\\$

Using initial value

$v(0)=\frac{1}{2}\\\\\frac{1}{2}=\frac{1}{0+2} +C\\\\C=0\\\\\frac{ds}{dt}=\frac{1}{t+2}$

We have to calculate the distance covered in time interval [0,5], so:

$\int\limits^5_0 \frac{ds}{dt}=\int\limits^5_0 {\frac{1}{t+2}} \, dt\\\\s(t)=[ln|t+2|]^5_0\\\\s(t)=ln|5+2|+ln|0+2|\\\\s(t)=(ln|7|+ln|2|)\,ft$

3 0

3 years ago

I need help ok I need to find the length of ab

Leokris [45]

I think you forgot to add a picture

6 0

3 years ago

If w=t and t=9,what is the value of w

finlep [7]

Answer: 9

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers