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2 years ago

PLEASE HELP ME!! I need you to solve this inequality. 2/3x - 1/5 > 1

Mathematics

1 answer:

Verizon [17]2 years ago

7 0

Answer:

If you are trying to solve the inequality for "x" Then your answer is, "x>9/5" That is the inequality form. If you are looking for the interval notation form then your answer is "9/5, infinity"

Step-by-step explanation:

I have checked my answers against an online Algebra solver. So my answers are 100% correct.

PLEASE MARK BRAINLIEST

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The position function of a particle in rectilinear motion is given by s(t) = 2t3 – 21t2 + 60t + 3 for t ≥ 0 with t measured in s

Norma-Jean [14]

The positions when the particle reverses direction are:

$s(t_1)=55ft\\\\s(t_2)=28ft$

The acceleraton of the paticle when reverses direction is:

$a(t_1)=-18\frac{ft}{s^{2}}\\ \\a(t_2)=a(5s)=18\frac{ft}{s^{2}}$

Why?

To solve the problem, we need to remember that if we derivate the position function, we will get the velocity function, and if we derivate the velocity function, we will get the acceleration function. So, we will need to derivate two times.

Also, when the particle reverses its direction, the velocity is equal to 0.

We are given the following function:

$s(t)=2t^{3}-21t^{2}+60t+3$

So,

- Derivating to get the velocity function, we have:

$v(t)=\frac{ds}{dt}=(2t^{3}-21t^{2}+60t+3)\\\\v(t)=3*2t^{2}-2*21t+60*1+0\\\\v(t)=6t^{2}-42t+60$

Now, making the function equal to 0, to find the times when the particle reversed its direction, we have:

$v(t)=6t^{2}-42t+60\\\\0=6t^{2}-42t+60\\\\0=t^{2}-7t+10\\(t-5)*(t-2)=0\\\\t_{1}=5s\\t_{2}=2s$

We know that the particle reversed its direction two times.

- Derivating the velocity function to find the acceleration function, we have:

$a(t)=\frac{dv}{dt}=6t^{2}-42t+60\\\\a(t)=12t-42$

Now, substituting the times to calculate the accelerations, we have:

$a(t_1)=a(2s)=12*2-42=-18\frac{ft}{s^{2}}\\ \\a(t_2)=a(5s)=12*5-42=18\frac{ft}{s^{2}}$

Now, substitutitng the times to calculate the positions, we have:

$s(t_1)=2*(2)^{3}-21*(2)^{2}+60*2+3=16-84+120+3=55ft\\\\s(t_2)=2*(5)^{3}-21*(5)^{2}+60*5+3=250-525+300+3=28ft$

Have a nice day!

3 0

3 years ago

Graph the line with slope 4 and y-intercept -9

Monica [59]

go down to (0, -9) on the graph, from there go up four and over one and plot the next point, up four over one again, and repeat

7 0

3 years ago

Hi, Could anyone help me with my homework

tatiyna

Answer:

$\hookrightarrow \sf x^6+24x^5+240x^4+1280x^3+3840x^2+6144x+4096$

solving steps:

$\rightarrow \sf (x + 4)^6$

$\bold{rewrite \ the \ following}$

$\rightarrow \sf (x + 4)^2 (x + 4)^2 (x + 4)^2$

$\bold {formula \ used : \sf (x+a)^2 = (x^2 + 2xa + a^2)}$

$\rightarrow \sf (x^2 + 8x+16) (x^2 + 8x+16) (x^2 + 8x+16)$

$\bold{simplify \ by \ removing \ parenthesis}$

$\rightarrow \sf (x^4 +8x^3 + 16x^2 + 8x^3 +64x^2 + 128x+16x^2+128x+256 ) (x^2 + 8x+16)$

$\bold{basic \ addition \ of \ integers }$

$\rightarrow \sf (x^4+16x^3+96x^2+256x+256) (x^2 + 8x+16)$

$\bold{remove \ parenthesis}$

$\rightarrow \sf (x^6 + 16x^5 + 96x^4 + 256x^3 + 256x^2 + 8x^6 + 128x^4 + 768x^3 + 2048x^2 + 2048 + 16x^4 + 256x^3 + 1536x^2 + 4096x + 4096)$

$\bold {final \ answer:}$

$\rightarrow \sf x^6+24x^5+240x^4+1280x^3+3840x^2+6144x+4096$

8 0

2 years ago

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denis-greek [22]

Answer:

4 is the median to your problem

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3 years ago

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OlgaM077 [116]

Answer:

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Step-by-step explanation:

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6 0

3 years ago