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3 years ago

Solve the equation fast please! I will say thank you and mark you Brainliest!

Mathematics

1 answer:

Mariulka [41]3 years ago

6 0

Answer:

x = ±5

Step-by-step explanation:

Multiply by (x-1)(x+1) and you get ...

(x+1) - (x -1) = (x -1)(x +1)/12

24 = x^2 -1 . . . . simplify and multiply by 12

0 = x^2 -25 . . . subtract 24

0 = (x -5)(x +5) . . . factor the difference of squares

Values of x that make the factors 0 are x=5 and x=-5.

The solutions are x=-5 or x=5.

_____

Subtracting the right side gives an equation that has its solutions at the x-intercepts of the left-side expression. These are seen to be -5 and +5 in the attached graph.

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Billy has a gallon of paint. He us going to pour it into a paint tray that measures 10 inches wide, 14 inches long and 4cm deep.

erica [24]

Answer:

Tray will overflow with $10.53\ \text{inch}^3$ of paint.

Step-by-step explanation:

Dimensions the tray is 10 inch by 14 inch by 4 cm

$1\ \text{cm}=\dfrac{1}{2.54}\ \text{inch}$

$4\ \text{cm}=\dfrac{4}{2.54}\ \text{inch}$

Volume the tray can hold is

$10\times 14\times \dfrac{4}{2.54}\ \text{in}^3$

$1\ \text{inch}^3=\dfrac{1}{231}\ \text{gallon}$

The volume of paint Billy has is $1\ \text{gallon}$

Difference in the volume of paint and volume of tray in cubic inches is

$\left(1-\dfrac{10\times14\times\dfrac{4}{2.54}}{231}\right)231=10.53\ \text{inch}^3$

The tray will overflow with $10.53\ \text{inch}^3$ of paint.

3 0

3 years ago

SIMPLE MATH PROBLEM! PLEASE HELP! The equation needs to use c, t, and =. Thanks!

Liula [17]

The cost c, is 55 per ticket t, so the equation is c=55t

6 0

3 years ago

Which of the following best describes the number shown below?

natita [175]

Answer:

Step-by-step explanation:

0.898989..... is a rational number because it has a terminating decimals

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3 years ago

A group of students are planning a mural at a wall the rectangular wall has dimensions of (6x+7) by (8x+5) and they are planning

zvonat [6]

Answer: $46x^2+73x+15$

Step-by-step explanation:

The area of a rectangle can be calculated with the formula:

$A=lw$

l: the length of the rectangle.

w: the width of the rectangle.

The area of the remaning wall after the mural has been painted, will be the difference of the area of the wall and the area of the mural.

Knowing that the dimensions of the wall are $(6x+7)$ by $(8x+5)$ , its area is:

$A_w=(6x+7)(8x+5)\\\\A_w=48x^2+30x+56x+35\\\\A_w=48x^2+86x+35$

As they are planning that the dimensions of the mural be $(x+4)$ by $(2x+5)$ , its area is:

$A_m=(x+4)(2x+5)\\\\A_m=2x^2+5x+8x+20\\\\A_m=2x^2+13x+20$

Then the area of the remaining wall after the mural has been painted is:

$A_{(remaining)}=A_w-A_m\\\\A_{(remaining)}=48x^2+86x+35-(2x^2+13x+20)\\\\A_{(remaining)}=48x^2+86x+35-2x^2-13x-20\\\\A_{(remaining)}=46x^2+73x+15$

8 0

3 years ago

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 x = 3 tan(t ) and dx = 3 sec²(t ) dt :

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

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

(b) The series

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

converges by comparison to the convergent p-series,

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

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

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 (n ² + 9)/eⁿ converges absolutely because ∑ |(-1)ⁿ (n ² + 9)/eⁿ| = ∑ (n ² + 9)/eⁿ in turn converges by comparison to a geometric series.

5 0

3 years ago