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2 years ago

Divide. −4 1/3÷−2 3/5 1 2/3 1 4/5 2 1/5 2 3/5

Mathematics

1 answer:

Aliun [14]2 years ago

6 0

Answer: 1 2/3

Step-by-step explanation:

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A. 40, B.80, C.120, D. 160, or E. 400

Rudik [331]

B: 80

All the sides add up to 20 so adding all four sides make 80

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2 years ago

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Help me I'm stuck on this problem

WITCHER [35]

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2 years ago

If I roll a standard number cube (labeled 1 to 6), and I roll it twice, what is the probability that I roll an odd number each t

suter [353]

Answer:

It is a 50% chance that you will roll and odd number each time.

Step-by-step explanation:

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3 years ago

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Which statement describes if there is an extraneous solution to the equation √x-3 = x-5? A. there are no solutions to the equati

Brrunno [24]

Remember that an extraneous solution of an equation, is the solution that emerges from solving the equation but is not a valid solution.

Lets solve our equation to find out what is the extraneous solution:
 $\sqrt{x-3} =x-5$
$(\sqrt{x-3})^2 =(x-5)^2$
$x-3=x^2-10x+25$
$x^2-11x+28=0$
$(x-4)(x-7)=0$
$x-4=0$ and $x-7=0$
$x=4$ and $x=7$

So, the solutions of our equation are $x=4$ and $x=7$ . Lets replace each solution in our original equation to check if they are valid solutions:
- For $x=7$
$\sqrt{x-3} =x-5$
$\sqrt{7-3} =7-5$
$\sqrt{4} =2$
$2=2$
We can conclude that 7 is a valid solution of the equation.

- For $x=4$
$\sqrt{x-3} =x-5$
$\sqrt{4-3} =4-5$
$\sqrt{1} =1$
$1 \neq 1$
We can conclude that 4 is not a valid solution of the equation; therefore, 4 is a extraneous solution.

We can conclude that the correct answer is: D. the extraneous solution is x = 4

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2 years ago

ASAP help due today brainliest

umka2103 [35]

B i think.. the product kinda gives it off to me but i’m not sure

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3 years ago

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