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Svetllana [295]
3 years ago
13

Solve the following inequalities 7 x minus 5 / 8 x + 3 >4​

Mathematics
2 answers:
olchik [2.2K]3 years ago
7 0

Answer:

x> \frac{8}{51}

Step-by-step explanation:

7x -  \frac{5}{8} x + 3>4

Bring constants to one side, simplify:

\frac{51}{8} x>4 - 3 \\  \frac{51}{8} x>1 \\ x>1 \div  \frac{51}{8}  \\ x>1 \times  \frac{8}{51}  \\ x> \frac{8}{51}

*Note that the inequality sign only changes when you divide the whole inequality by a negative number.

GREYUIT [131]3 years ago
7 0

Answer:

x>\frac{8}{51}

Step-by-step explanation:

7x-\frac{5}{8}x+3>4\\\mathrm{Subtract\:}3\mathrm{\:from\:both\:sides}\\7x-\frac{5}{8}x+3-3>4-3\\\mathrm{Simplify}\\7x-\frac{5}{8}x>1\\\mathrm{Multiply\:both\:sides\:by\:}8\\7x\times \:8-\frac{5}{8}x\times \:8>1\times \:8\\\mathrm{Simplify}\\56x-5x>8\\51x>8\\\mathrm{Divide\:both\:sides\:by\:}51\\\frac{51x}{51}>\frac{8}{51}\\\\x>\frac{8}{51}

I hope it helps :)

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3 years ago
Subtract 7x-9 from2x^2-11
strojnjashka [21]

Answer:

The answer is: 2x^2-7x-2

Step-by-step explanation:

we are asked to subtract 7x-9 from 2x^2-11

so, 2x^2-11-(7x-9)\\=2x^2-11-7x+9\\=2x^2-7x-2.

Hence the desired result is:  2x^2-7x-2.


8 0
3 years ago
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Complete parts a through c for the given function.
victus00 [196]

Answer:

a. The critcal points are at

x=0,-5,3

b. Then, x = -5   is a maximum and x=3 is a minimum

c. The absolute minimum is at   x = 3  and the absolute maximum is at  x = -5.

Step-by-step explanation:

(a)

Remember that you need to find the points where

f'(x)=0

Therefore you have to solve this equation.

20x^4  + 40x^3 - 300x^2 = 0

From that equation you can factor out    20x^2  and you would get

20x^2 (  x^2  +2x - 15)  = 0

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And you would also have  x^2 +2x-15 = 0.

You can factor that equation as    x^2 +2x -15 = (x+5)(x-3) = 0

Therefore   x=-5 ,   x=3.

So the critcal points are at

x=0,-5,3

b.  

Remember that a function has a maximum at a critical point if the second derivative at that point is negative. Since

f''(x) = 80x^3 + 120x^2 -600x\\f''(-5) = 80(-5)^3 + 120(-5)^2 -600(-5) = -4000 < 0\\\\f''(3) = 80(3)^3 + 120(3)^2 -600(3) = 1440 > 0 \\

Then, x = -5   is a maximum and x=3 is a minimum

c.

The absolute minimum is at   x = 3  and the absolute maximum is at  x = -5.

6 0
3 years ago
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Square root of 512m^3. show work
ElenaW [278]

The solution would be like this for this specific problem:

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I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

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