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DerKrebs [107]
3 years ago
6

Over and 12 hour period from 8 p.m. to 8 a.m. the temperature fell at a steady rate of 8 f to -16 f. if the temperature fell at

the same rate every hour, what was the temperature at 4 a.m.
Mathematics
1 answer:
Kamila [148]3 years ago
7 0
The steady drop rate of the temperature is 2, 8 p.m. to 4 a.m. is 7 hours. that would make the temperature at 4 a.m. is -6 f.
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abruzzese [7]

The quadratic formula is:

\frac{-b + \sqrt{b^2 - 4ac} }{2a} and  \frac{-b - \sqrt{b^2 - 4ac} }{2a}

In this case, a = 2, b = -3, and c = -7

So, we can plug in the numbers to get:

  \frac{-(-3) + \sqrt{(-3)^2 - 4(2)(-7)} }{2(2)} and    \frac{-(-3) - \sqrt{(-3)^2 - 4(2)(-7)} }{2(2)}

Simplifying, we get:

    \frac{3 + \sqrt{65} }{4} and   \frac{3 - \sqrt{65} }{4}

You need to use a calculator to find what these would be in decimal form. The answer, rounded to the nearest tenth, is: x = 2.8, x = -1.3

8 0
3 years ago
Find+the+positive+value+for+α+if+the+radius+of+the+circle+3x^2+3y-6αx+12y-3α=0+is+4
Alik [6]

Answer:

\alpha=3

Step-by-step explanation:

<u>Equation of a Circle</u>

A circle of radius r and centered on the point (h,k) can be expressed by the equation

(x-h)^2+(y-k)^2=r^2

We are given the equation of a circle as

3x^2+3y^2-6\alpha x+12y-3\alpha=0

Note we have corrected it by adding the square to the y. Simplify by 3

x^2+y^2-2\alpha x+4y-\alpha=0

Complete squares and rearrange:

x^2-2\alpha x+y^2+4y=\alpha

x^2-2\alpha x+\alpha^2+y^2+4y+4=\alpha+\alpha^2+4

(x-\alpha)^2+(y+2)^2=r^2

We can see that, if r=4, then

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There are two solutions for \alpha:

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Keeping the positive solution, as required:

\boxed{\alpha=3}

8 0
3 years ago
a triangle has area 49.5 cm^2. if the base of the triangle is 9 cm, what is the height of the triangle
Anika [276]
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height = (Area * 2) / base
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3 0
3 years ago
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ser-zykov [4K]

Answer:

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Step-by-step explanation:

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Substitute t=0

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Substitute the values

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5 0
3 years ago
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Anton [14]

Answer:

Step-by-step explanation:

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