PLS HELP ME!! FIND X!!

Mathematics

1 answer:

bija089 [108]3 years ago

6 0

Answer:

I got 1

Step-by-step explanation:

I dont know if I go it right

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In 2-3 sentences describe why the only way an infinite geometric series has a sum is if |r| < 1.

Lina20 [59]

The formula for the sum of geometric series = a (1 - r^n) / 1 - r
when it will be an infinite geometric series. the term r^n will equal to zero because and number < 1 that is raise to the power infinity the answer is zero that is why in a infinite geometric series, sum = a / 1 - r.
if the r < 1 then there will be no numerical answer because the sum will be infinity. that is why r < 1 to have a sum

4 0

3 years ago

Inequalities <br><img src="https://tex.z-dn.net/?f=%20-%205%20%5Ctimes%20%20%2B%202%20%5Cgeqslant%20%20-%2048" id="TexFormula1"

Andrej [43]

$\bf -5x+2\geqslant -48\implies -5x \geqslant -50\implies \stackrel{\textit{now, we're dividing by -5}}{\cfrac{-5x}{-5}\stackrel{\stackrel{\downarrow }{\frac{}{}}}{\leqslant}\cfrac{-50}{-5}}\implies x\leqslant 10$

5 0

3 years ago

Find the area of each sector. Show your work.

hodyreva [135]

Area of the shaded sector = (120/360) * pi * 10^2

= 104.72 square units

Area of the other sector = 2 * 104.72 = 209.44 sq units

6 0

3 years ago

Evaluate the integral by interpreting it in terms of areas Draw a picture of the region the integral

denis23 [38]

Answer:

Step-by-step explanation:

The picture is below of how to separate this into 2 different regions, which you have to because it's not continuous over the whole function. It "breaks" at x = 2. So the way to separate this is to take the integral from x = 0 to x = 2 and then add it to the integral for x = 2 to x = 3. In order to integrate each one of those "parts" of that absolute value function we have to determine the equation for each line that makes up that part.

For the integral from [0, 2], the equation of the line is -3x + 6;

For the integral from [2, 3], the equation of the line is 3x - 6.

We integrate then:

$\int\limits^2_0 {-3x+6} \, dx+\int\limits^3_2 {3x-6} \, dx$ and

$-\frac{3x^2}{2}+6x\left \} {{2} \atop {0}} \right. +\frac{3x^2}{2}-6x\left \} {{3} \atop {2}} \right.$ sorry for the odd representation; that's as good as it gets here!