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tino4ka555 [31]
3 years ago
15

A physics lab group was conducting an experiment to determine the length of a spring when different objects of varying weight we

re hung from it. After testing ten different objects, the group calculated the following linear regression where x is the weight of the object in ounces and h(x) is the length of the spring in centimeters: h(x)= 2.3x+15.5. What does y-intercept of this equation indicate about the relationship between object weight and the length of the spring?
1) When there is no object on the spring, its length is 2.3 cm
2) When there is no object on the spring, its length 15.5 cm
3) For every ounce of weight, the spring length increases by 2.3 cm
4) For every ounce of weight, the spring length increases by 1535 cm
Mathematics
2 answers:
Mila [183]3 years ago
5 0
2). The length of the spring is 15.5 cm, and that's how it hangs
when there's no weight on it.

In the equation, 'x' is the weight of the object. If there's no object, then
x = 0 and h(x) = 15.5 .
Lemur [1.5K]3 years ago
3 0
15.5 is the answer
man...that was pretty hard
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rusak2 [61]

Answer:

The zeros are

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Step-by-step explanation:

We have been given the equation x^4-6x^2-7x-6=0

Use rational root theorem, we have

a_0=6,\:\quad a_n=1

\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:2,\:3,\:6,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1

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=\left(x+2\right)\frac{x^4-6x^2-7x-6}{x+2}\\

=x^3-2x^2-2x-3

Again factor using the rational root test, we get

=\left(x+2\right)\left(x-3\right)\left(x^2+x+1\right)

Using the zero product rule, we have

x+2=0:\quad x=-2\\x-3=0:\quad x=3\\x^2+x+1=0\\\\x_{1,\:2}=\frac{-1\pm \sqrt{1^2-4\cdot \:1\cdot \:1}}{2\cdot \:1}\\\\=-\frac{1}{2}+i\frac{\sqrt{3}}{2},\:x=-\frac{1}{2}-i\frac{\sqrt{3}}{2}

Therefore, the zeros are

x=-2,\:x=3,\:x=-\frac{1}{2}+i\frac{\sqrt{3}}{2},\:x=-\frac{1}{2}-i\frac{\sqrt{3}}{2}


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