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nignag [31]
3 years ago
14

What are the real or imaginary solutions of the polynomial equation? x^3-8=0

Mathematics
1 answer:
Flura [38]3 years ago
8 0

here,

x³-8=0

x³=8

x³=2³

x=2

I hope this may help you

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alina1380 [7]
-1/6 thats the answer
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To raise money for a local charity, students organized a wake-a-thon
arsen [322]

Answer:

a) T= 3.125h

b) $75

Step-by-step explanation:

Given: We know Tania raised $50 for staying awake 16 hours. We also know the amount of money raised varies directly with the time.

We can divide $50 into 16 hours.

50/16 = 25/8

25/8 = 3.125

3.125 = 3 1/8 (You can use a fraction or decimal, whichever is preferred)

That's how much money she earns per hour. It may not make sense because you have 1/8 of a dollar, so it's okay if you're a little confused on that.

Now let's answer the first question: Write an equation relating the money Tania raised and the  amount of time, in hours, she stayed awake.

We know Tania earns $3.125 per hour, so we can make it this equation

T= 3.125h

T can represent Tania's earnings, and 3.125 is how much she earns h, hourly.

Now onto the second question: How much would she have raised by staying awake for 24 h?

We need to multiply 3.125, or 3 1/8, 24 times. Your answer should be $75.

(I apologize in advance if I misread something or made a mistake)

6 0
3 years ago
A line passes through (10,4) and (13,-11). Write the equation of the line in standard form.
AlexFokin [52]

\bf (\stackrel{x_1}{10}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{13}~,~\stackrel{y_2}{-11}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-11-4}{13-10}\implies \cfrac{-15}{3}\implies -5 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-4=-5(x-10)


now

standard form for a linear equation means

• all coefficients must be integers, no fractions

• only the constant on the right-hand-side

• all variables on the left-hand-side, sorted

• "x" must not have a negative coefficient

therefore,


\bf y-4=-5(x-10)\implies y-4=-5x+50\implies \blacktriangleright 5x+y=54 \blacktriangleleft

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