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musickatia [10]
3 years ago
11

Write an equation in point-slope form for the line that has the given slope and that contains the given point. slope 4/5 ,(8,2)

Mathematics
1 answer:
Novosadov [1.4K]3 years ago
6 0
Hello,

The equation is y-2=4/5(x-8)==>y=4/5x-22/5 or 4x-5y=22
=========================================
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2.5 x 2.5 x 2.5 - 1.4 x 1.4 x 1.4<br>(2.5)2+ 2.5 X 1.4 + (1.4)2​
melomori [17]

Answer:

a) 2.5×2.5×2.5-1.4×1.4×1.4=12.881

b) (2.5)2+2.5×1.4+(1.4)2=31.5

Step-by-step explanation:

yuh multiple 2.5 three times and subtract 1.4 multiple three times. 2.5×2.5×2.5-1.4×1.4×1.4

15.625-2.744=12.881.

b) when opening bracket Yuh multiple the number in the bracket by the number outside the bracket. when bracket is open Yuh can now add, multiple to obtain your answer please

(2.5)2+2.5×1.4+(1.4)2

5+2.5×1.4+2.8

7.5×4.2=31.5

8 0
3 years ago
Please help po!!! ASAP
zalisa [80]

Answer:

I drawing nyo po tapos Yun na madali Lang yan

6 0
3 years ago
PLEASE HELP ASAP
Dafna1 [17]

Answer:

Step-by-step explanation:

27.068

28.08

28.15

29.94

6 0
2 years ago
Read 2 more answers
Let f be defined by the function f(x) = 1/(x^2+9)
riadik2000 [5.3K]

(a)

\displaystyle\int_3^\infty \frac{\mathrm dx}{x^2+9}=\lim_{b\to\infty}\int_{x=3}^{x=b}\frac{\mathrm dx}{x^2+9}

Substitute <em>x</em> = 3 tan(<em>t</em> ) and d<em>x</em> = 3 sec²(<em>t </em>) d<em>t</em> :

\displaystyle\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\frac{3\sec^2(t)}{(3\tan(t))^2+9}\,\mathrm dt=\frac13\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\mathrm dt

=\displaystyle \frac13 \lim_{b\to\infty}\left(\arctan\left(\frac b3\right)-\arctan(1)\right)=\boxed{\dfrac\pi{12}}

(b) The series

\displaystyle \sum_{n=3}^\infty \frac1{n^2+9}

converges by comparison to the convergent <em>p</em>-series,

\displaystyle\sum_{n=3}^\infty\frac1{n^2}

(c) The series

\displaystyle \sum_{n=1}^\infty \frac{(-1)^n (n^2+9)}{e^n}

converges absolutely, since

\displaystyle \sum_{n=1}^\infty \left|\frac{(-1)^n (n^2+9)}{e^n}\right|=\sum_{n=1}^\infty \frac{n^2+9}{e^n} < \sum_{n=1}^\infty \frac{n^2}{e^n} < \sum_{n=1}^\infty \frac1{e^n}=\frac1{e-1}

That is, ∑ (-1)ⁿ (<em>n</em> ² + 9)/<em>e</em>ⁿ converges absolutely because ∑ |(-1)ⁿ (<em>n</em> ² + 9)/<em>e</em>ⁿ| = ∑ (<em>n</em> ² + 9)/<em>e</em>ⁿ in turn converges by comparison to a geometric series.

5 0
3 years ago
Proportion problem<br><br><br><img src="https://tex.z-dn.net/?f=%5Cfrac%7B8%7D%7B9%7D" id="TexFormula1" title="\frac{8}{9}" alt=
kipiarov [429]

Answer: w=66/9

Step-by-step explanation: 8/9=w-2/6

8(6)=9(w-2)

48 =9w-18

48+18=9w

66=9w

w=66/9

or w=7(3/9)

6 0
3 years ago
Read 2 more answers
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