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Three ways to write 3 to the 5th power

cestrela7 [59]

First expression

=> 3 x 3 x 3 x 3 x 3 – In this expression, we multiplied 3 5 times and the product is 243

Second expression

=> 3 ^ 5 , in where ^ read as raised to the power. , the product is also 243

Third expression

=> 3^2 x 3 ^3

=> (3 x 3) x (3 x 3 x 3)

=> 9 x 27, the product is also equals to 243.

7 0

3 years ago

Read 2 more answers

I am in a algebra 1 class and we are learning about variables and slope and all that stuff , and also were mainly talking about

kozerog [31]

Answer:

okay .. then what is your questions ..

8 0

2 years ago

Determine the singular points of the given differential equation. Classify each singular point as regular or irregular. (Enter y

ludmilkaskok [199]

Answer:

Step-by-step explanation:

Given that:

The differential equation; $(x^2-4)^2y'' + (x + 2)y' + 7y = 0$

The above equation can be better expressed as:

$y'' + \dfrac{(x+2)}{(x^2-4)^2} \ y'+ \dfrac{7}{(x^2- 4)^2} \ y=0$

The pattern of the normalized differential equation can be represented as:

y'' + p(x)y' + q(x) y = 0

This implies that:

$p(x) = \dfrac{(x+2)}{(x^2-4)^2} \$

$p(x) = \dfrac{(x+2)}{(x+2)^2 (x-2)^2} \$

$p(x) = \dfrac{1}{(x+2)(x-2)^2}$

Also;

$q(x) = \dfrac{7}{(x^2-4)^2}$

$q(x) = \dfrac{7}{(x+2)^2(x-2)^2}$

From p(x) and q(x); we will realize that the zeroes of (x+2)(x-2)² = ±2

When x = - 2

$\lim \limits_{x \to-2} (x+ 2) p(x) = \lim \limits_{x \to2} (x+ 2) \dfrac{1}{(x+2)(x-2)^2}$

$\implies \lim \limits_{x \to2} \dfrac{1}{(x-2)^2}$

$\implies \dfrac{1}{16}$

$\lim \limits_{x \to-2} (x+ 2)^2 q(x) = \lim \limits_{x \to2} (x+ 2)^2 \dfrac{7}{(x+2)^2(x-2)^2}$

$\implies \lim \limits_{x \to2} \dfrac{7}{(x-2)^2}$

$\implies \dfrac{7}{16}$

Hence, one (1) of them is non-analytical at x = 2.

Thus, x = 2 is an irregular singular point.

5 0

3 years ago

Determine if the following system of equations has no solutions 4x+3y= -8 -8x-6y= 16

cupoosta [38]

$\begin{cases} 4x+3y=-8\\\\ -8x-6y=16 \end{cases}~\hspace{10em} \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array} \\\\[-0.35em] ~\dotfill$

$4x+3y=-8\implies 3y=-4x-8\implies y=\cfrac{-4x-8}{3}\implies y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{4}{3}} x-\cfrac{8}{3} \\\\[-0.35em] ~\dotfill\\\\ -8x-6y=16\implies -6y=8x+16\implies y=\cfrac{8x+16}{-6} \\\\\\ y=\cfrac{8}{-6}x+\cfrac{16}{-6}\implies y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{4}{3}} x-\cfrac{8}{3}$

one simple way to tell if both equations do ever meet or have a solution is by checking their slope, notice in this case the slopes are the same for both, meaning the lines are parallel lines, however, notice both equations are really the same, namely the 2nd equation is really the 1st one in disguise.

since both equations are equal, their graph will be of one line pancaked on top of the other, and the solutions is where they meet, hell, they meet everywhere since one is on top of the other, so infinitely many solutions.

3 0

2 years ago

Mia made three times as many baskets in yesterday’s basketball game as she did last Saturday. She scored a total of 24 baskets f

natima [27]

Answer:

8

Step-by-step explanation:

i did 24 divided by 3

hope this helps

-mercury

5 0

3 years ago