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user100 [1]
3 years ago
5

Enter an inequality that represents the description, and then solve.

Mathematics
1 answer:
natta225 [31]3 years ago
5 0

Answer:

X >= 7

Step-by-step explanation:

7x + 8 >= 57

7x >= 57-8

7x >= 49

x >= 49/7

x >= 7

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0.6x 4.8=7.2 without integers
velikii [3]
2.88x=7.2
x=7.2/2.88
x=2.5
5 0
3 years ago
how can you recognize whether a real-world situation should be represented by an equation or an inequality
zaharov [31]
Real-world situations can be represented by an equation when something is strictly equal to another thing (ex. when the price of a stick of gum is the price as two erasers) Real-world situations can be represented by inequalities when something is not defitively equal to another (ex. say that Sam wanted to buy snacks and water for her soccer team, she can spend no more than $20, here you would use an inequality symbol)

It might be a bit complicated to understand my wording in its entirety, but I hope that I helped!!
6 0
3 years ago
If PQR measures 75°, what is the measure of SQR?<br><br> 22°<br> 45°<br> 53°<br> 97°
allochka39001 [22]
C is the right answer
4 0
3 years ago
Read 2 more answers
Find all the zeroes of the equation. -3x4+27x2+1200=0 I need to learn how to do it! Just not the answer please
Mumz [18]
The first thing you should do for this case is to rewrite the expression.
 To do this, divide both sides of the equation between -3.
 We have then:
 (-1/3) * (- 3x ^ 4 + 27x ^ 2 + 1200) = (- 1/3) * (0)
 x ^ 4-9x ^ 2-400 = 0
 Then, we rewrite the polynomial:
 (x-5) * (x + 5) * (x ^ 2 + 16) = 0
 From here, we obtain the four roots:
 Root 1:
 x-5 = 0
 x = 5
 Root 2:
 x + 5 = 0
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 Root 3 and root 4:
 x ^ 2 + 16 = 0
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 x4 = 4i

 Answer:
 all the zeroes of the equation are: 
 x1 = 5 
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x4 = 4i
4 0
4 years ago
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Will mark as brainliest :) <br> Simplify the Difference.
kotykmax [81]

Answer:

\frac{-2x \times (x+3)}{(x-2)(x+1)(x-1)}

Step-by-step explanation:

we are given

\frac{2x}{x^2-x-2} - \frac{4x}{x^2-3x+2}           ----------(A)

Let us simplify the two denominators first. One by one

x^2-x-2

= x^2-2x+x-2

= x(x-2)-1(x-2)

= (x-2)(x+1)

{x^2-3x+2}

=x^2-x-2x+2

=x(x-1)-2(x-1)

=(x-1)(x-2)

Hence (A) becomes

\frac{2x}{x^2-x-2} - \frac{4x}{x^2-3x+2}

= \frac{2x}{(x-2)(x+1)} - \frac{4x}{(x-2)(x-1)}

taking out \frac{2x}{x-2}  as GCD

\frac{2x}{x-2}( \frac{1}{x+1} - \frac{2}{x-1}

\frac{2x}{x-2}(\frac{(x-1)-2(x+1)}{(x+1)(x-1)}

\frac{2x}{x-2}(\frac{x-1-2x-2)}{(x+1)(x-1)}

\frac{2x}{x-2}(\frac{(-x-3)}{(x+1)(x-1)}

\frac{-2x \times (x+3)}{(x-2)(x+1)(x-1)}

7 0
3 years ago
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