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babymother [125]
3 years ago
9

Is 13 a perfect cube , perfect square , or neither

Mathematics
2 answers:
denis-greek [22]3 years ago
5 0

Answer:

neither

Step-by-step explanation:

if we cube root or square root thirteen, the result will be in decimal. So thirteen is neither a perfect square nor a perfect cube

Please mark as brainliest

True [87]3 years ago
5 0

nope, neither.

Scroll up and give the man branliest, he asked first.

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Answer:

First option:  3x^2-5x=-8

Second option: 2x^2=6x-5

Fourth option: -x^2-10x=34

Step-by-step explanation:

Rewrite each equation in the form ax^2+bx+c=0 and then use the Discriminant formula for each equation. This is:

D=b^2-4ac

1) For 3x^2-5x=-8:

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Then:

D=(-5)^2-4(3)(8)=-71

Since D this equation has no real solutions, but has two complex solutions.

2) For 2x^2=6x-5:

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Then:

D=(-6)^2-4(2)(5)=-4

Since D this equation has no real solutions, but has two complex solutions.

3) For 12x=9x^2+4:

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Then:

D=(-12)^2-4(9)(4)=0

Since D=0 this equation has one real solution.

4) For -x^2-10x=34:

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Then:

D=(-10)^2-4(-1)(-34)=-36

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3 years ago
Solve for y , what is 9x+3y=4 ??
lora16 [44]
To solve for y, isolate the variable to one side.
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~~~~~~~~
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4 years ago
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MA_775_DIABLO [31]

Answer:

y = 5 e^r * t

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y = 5 e^(.0224 t)    is then our equation

Check  - suppose you want y at 2020

y = 5 e^(.0224 * 20)    would be the equation

y = 5 e^.449 = 7.83  billion - seems to be a reasonable answer

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4.8541019662497 + 1.8541019662497 = -3

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PLEASE HELP ASAP!!! A talent competition on television had five elimination rounds. After each elimination, only one-third of th
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