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Ksivusya [100]
2 years ago
5

How can Ari simplify the following expression? StartFraction 5 Over a minus 3 EndFraction minus 4 divided by 2 + StartFraction 1

Over a minus 3 EndFraction
Mathematics
1 answer:
Zielflug [23.3K]2 years ago
8 0

Answer:

-\frac{8}{3}

Step-by-step explanation:

Given

\frac{5}{-3} - \frac{4}{2} + \frac{1}{-3}

Required

Simpify

The very first step is to take LCM of the given expression

\frac{-10 -4 - 2}{6}

Perform arithmetic operations o the numerator

-\frac{16}{6}

Divide the numerator and denominator by 2

-\frac{16/2}{6/2}

-\frac{8}{3}

The expression can't be further simplified;

Hence, \frac{5}{-3} - \frac{4}{2} + \frac{1}{-3} = -\frac{8}{3}

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Lisa says that the indicated angles cannot have the same measure. Marita disagrees and says she can prove that they can have the
AlexFokin [52]
I agree with Marita, that the angles could have the same measure.  This can be proven if you set the two amounts equal and solve for x. 

9x - 25 + x = x + 50 + 2x - 12

To begin, we should combine like terms on both sides of the equation to start simplifying the equation.

10x - 25 = 3x + 38

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7x = 63

Finally, we should divide both sides by 7, to get rid of the coefficient of x.

x = 9

If you plug in 9 for x in our first equation, you get that both of the angle measurements equal 65 degrees.  This means that Marita is correct, because if x = 9, then the angles would have the same measure.


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2 years ago
Part 1: Graph piecewise functions from tables and equations. Isabel sells new and used mystery novels online. Her customers are
Charra [1.4K]
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2 years ago
Maria has 14 pencils, Ted has 9 pencils, Tom has 12 pencils, Lucinda has 13 pencils, and Jumana has 12 pencils. How can they red
Alisiya [41]

5 people.

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7 0
2 years ago
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The fish population in a pond with carrying capacity 1200 is modeled by the following logistic equation where N(t) denotes the n
MArishka [77]

9514 1404 393

Answer:

  (dN)/(dt) = (0.4)/(1200)N(1200-N) -50

  142 fish

Step-by-step explanation:

A) The differential equation is modified by adding a -50 fish per year constant term:

  (dN)/(dt) = (0.4)/(1200)N(1200-N) -50

__

B) The steady-state value of the fish population will be when N reaches the value that makes dN/dt = 0.

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This steady-state number of fish is ...

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8 0
2 years ago
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kherson [118]
I believe the maximum number of pounds would be 11.

8 0
2 years ago
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