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Ipatiy [6.2K]
3 years ago
14

The store Clothes-4-You keeps data on their sales. These data points represent their coat sales, where x is the average outside

temperature and y is the number of coats sold. (3, 94), (12, 91), (16, 83), (21, 93), (27, 82), (29, 69), (40, 71), (43, 63), (60, 38), (52, 53) What is the best estimate for the number of coats sold when the outside temperature is 35°F?
Mathematics
2 answers:
ikadub [295]3 years ago
5 0

Answer:

the answer is 70 coats

Step-by-step explanation:

if 35 is the temperature that you need to plug in the points on the graphs X and Y coordinators to find that the answer is 70

Alex3 years ago
4 0

Answer:

It should be 70 coats.

Step-by-step explanation:

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Answer:

8 [ (x-3) (2x+1) ]

Step-by-step explanation:

Factoring  2x2 - 5x - 3  

The first term is,  2x2  its coefficient is  2 .

The middle term is,  -5x  its coefficient is  -5 .

The last term, "the constant", is  -3  

Step-1 : Multiply the coefficient of the first term by the constant   2 • -3 = -6  

Step-2 : Find two factors of  -6  whose sum equals the coefficient of the middle term, which is   -5 .

     -6    +    1    =    -5    That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -6  and  1  

                    2x2 - 6x + 1x - 3

Step-4 : Add up the first 2 terms, pulling out like factors :

                   2x • (x-3)

             Add up the last 2 terms, pulling out common factors :

                    1 • (x-3)

Step-5 : Add up the four terms of step 4 :

                   (2x+1)  •  (x-3)

            Which is the desired factorization

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Factor 15x3-5x+6x-2 by grouping. What is the resulting expression?
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Step-by-step explanation:

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faltersainse [42]

Answer:

Procedure:

1) Form a system of 3 linear equations based on the two zeroes and a point.

2) Solve the resulting system by analytical methods.

3) Substitute all coefficients.

Step-by-step explanation:

A quadratic function is a polynomial of the form:

y = a\cdot x^{2}+b\cdot x + c (1)

Where:

x - Independent variable.

y - Dependent variable.

a, b, c - Coefficients.

A value of x is a zero of the quadratic function if and only if y = 0. By Fundamental Theorem of Algebra, quadratic functions with real coefficients may have two real solutions. We know the following three points: A(x,y) = (r_{1}, 0), B(x,y) = (r_{2},0) and C(x,y) = (x,y)

Based on such information, we form the following system of linear equations:

a\cdot r_{1}^{2}+b\cdot r_{1} + c = 0 (2)

a\cdot r_{2}^{2}+b\cdot r_{2} + c = 0 (3)

a\cdot x^{2} + b\cdot x + c = y (4)

There are several forms of solving the system of equations. We decide to solve for all coefficients by determinants:

a = \frac{\left|\begin{array}{ccc}0&r_{1}&1\\0&r_{2}&1\\y&x&1\end{array}\right| }{\left|\begin{array}{ccc}r_{1}^{2}&r_{1}&1\\r_{2}^{2}&r_{2}&1\\x^{2}&x&1\end{array}\right| }

a = \frac{y\cdot r_{1}-y\cdot r_{2}}{r_{1}^{2}\cdot r_{2}+r_{2}^{2}\cdot x+x^{2}\cdot r_{1}-x^{2}\cdot r_{2}-r_{2}^{2}\cdot r_{1}-r_{1}^{2}\cdot x}

a = \frac{y\cdot (r_{1}-r_{2})}{r_{1}^{2}\cdot r_{2}+r_{2}^{2}\cdot x +x^{2}\cdot r_{1}-x^{2}\cdot r_{2}-r_{2}^{2}\cdot r_{1}-r_{1}^{2}\cdot x}

b = \frac{\left|\begin{array}{ccc}r_{1}^{2}&0&1\\r_{2}^{2}&0&1\\x^{2}&y&1\end{array}\right| }{\left|\begin{array}{ccc}r_{1}^{2}&r_{1}&1\\r_{2}^{2}&r_{2}&1\\x^{2}&x&1\end{array}\right| }

b = \frac{(r_{2}^{2}-r_{1}^{2})\cdot y}{r_{1}^{2}\cdot r_{2}+r_{2}^{2}\cdot x +x^{2}\cdot r_{1}-x^{2}\cdot r_{2}-r_{2}^{2}\cdot r_{1}-r_{1}^{2}\cdot x}

c = \frac{\left|\begin{array}{ccc}r_{1}^{2}&r_{1}&0\\r_{2}^{2}&r_{2}&0\\x^{2}&x&y\end{array}\right| }{\left|\begin{array}{ccc}r_{1}^{2}&r_{1}&1\\r_{2}^{2}&r_{2}&1\\x^{2}&x&1\end{array}\right| }

c = \frac{(r_{1}^{2}\cdot r_{2}-r_{2}^{2}\cdot r_{1})\cdot y}{r_{1}^{2}\cdot r_{2}+r_{2}^{2}\cdot x + x^{2}\cdot r_{1}-x^{2}\cdot r_{2}-r_{2}^{2}\cdot r_{1}-r_{1}^{2}\cdot x}

And finally we obtain the equation of the quadratic function given two zeroes and a point.

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Answer:

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