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olga2289 [7]
3 years ago
7

Please help me please !

Mathematics
1 answer:
Ratling [72]3 years ago
6 0

9514 1404 393

Answer:

  Molly

Step-by-step explanation:

Brady did not properly create a square.

The square of a binomial is ...

  (a -b)² = a² -2ab +b²

Where a²=x² and -2ab = -4x, you find b=2. The constant term in the square is 2²=4. Brady shows it as -4, which is incorrect.

Molly's work is correct.

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Please answer this question​
Tems11 [23]

\bold{\huge{\underline{ Solution }}}

<h3><u>Given </u><u>:</u><u>-</u><u> </u></h3>

• \sf{ Polynomial :- ax^{2} + bx + c }

• The zeroes of the given polynomial are α and β .

<h3><u>Let's </u><u>Begin </u><u>:</u><u>-</u><u> </u></h3>

Here, we have polynomial

\sf{ = ax^{2} + bx + c }

<u>We </u><u>know </u><u>that</u><u>, </u>

Sum of the zeroes of the quadratic polynomial

\sf{ {\alpha} + {\beta} = {\dfrac{-b}{a}}}

<u>And </u>

Product of zeroes

\sf{ {\alpha}{\beta} = {\dfrac{c}{a}}}

<u>Now, we have to find the polynomials having zeroes </u><u>:</u><u>-</u>

\sf{ {\dfrac{{\alpha} + 1 }{{\beta}}} ,{\dfrac{{\beta} + 1 }{{\alpha}}}}

<u>T</u><u>h</u><u>erefore </u><u>,</u>

Sum of the zeroes

\sf{ ( {\alpha} + {\dfrac{1 }{{\beta}}} )+( {\beta}+{\dfrac{1 }{{\alpha}}})}

\sf{ ( {\alpha} + {\beta}) + ( {\dfrac{1}{{\beta}}} +{\dfrac{1 }{{\alpha}}})}

\sf{( {\dfrac{ -b}{a}} ) + {\dfrac{{\alpha}+{\beta}}{{\alpha}{\beta}}}}

\sf{( {\dfrac{ -b}{a}} ) + {\dfrac{-b/a}{c/a}}}

\sf{ {\dfrac{ -b}{a}} + {\dfrac{-b}{c}}}

\bold{{\dfrac{ -bc - ab}{ac}}}

Thus, The sum of the zeroes of the quadratic polynomial are -bc - ab/ac

<h3><u>Now</u><u>, </u></h3>

Product of zeroes

\sf{ ( {\alpha} + {\dfrac{1 }{{\beta}}} ){\times}( {\beta}+{\dfrac{1 }{{\alpha}}})}

\sf{ {\alpha}{\beta} + 1 + 1 + {\dfrac{1}{{\alpha}{\beta}}}}

\sf{ {\alpha}{\beta} + 2 + {\dfrac{1}{{\alpha}{\beta}}}}

\bold{ {\dfrac{c}{a}} + 2 + {\dfrac{ a}{c}}}

Hence, The product of the zeroes are c/a + a/c + 2 .

<u>We </u><u>know </u><u>that</u><u>, </u>

<u>For </u><u>any </u><u>quadratic </u><u>equation</u>

\sf{ x^{2} + ( sum\: of \:zeroes )x + product\:of\: zeroes }

\bold{ x^{2} + ( {\dfrac{ -bc - ab}{ac}} )x + {\dfrac{c}{a}} + 2 + {\dfrac{ a}{c}}}

Hence, The polynomial is x² + (-bc-ab/c)x + c/a + a/c + 2 .

<h3><u>Some </u><u>basic </u><u>information </u><u>:</u><u>-</u></h3>

• Polynomial is algebraic expression which contains coffiecients are variables.

• There are different types of polynomial like linear polynomial , quadratic polynomial , cubic polynomial etc.

• Quadratic polynomials are those polynomials which having highest power of degree as 2 .

• The general form of quadratic equation is ax² + bx + c.

• The quadratic equation can be solved by factorization method, quadratic formula or completing square method.

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2 years ago
Please help, this is a really big test. Please be correct as well :)<br> Thank you!
xenn [34]

The correct answer is: D (2,1)

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What dose equivalent mean
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EQUI = equals

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EQUI + VALent, same value

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(3+1/2) ×12. multiple or divide​
taurus [48]

Answer:

(3+1/2) x 12 = 42

Step-by-step explanation:

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3 years ago
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Write an equation showing the relationship between the number of baskets (b) made from outside the three-point line and the numb
Ghella [55]

The equation that shows the relationship between the number of baskets (b) made from outside the three-point line and the number of points scored (p) is p = 3b

<h3>How to find an equation using relationship of variables?</h3>

In basket ball, If a shot is successfully scored from outside of the three-point line, three points are awarded.

where

  • p = number of points scored
  • b = number of basket made outside the three point line.

Therefore, the equation that shows the relationship between the number of baskets (b) made from outside the three-point line and the number of points scored (p) is as follows:

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learn more equation here: brainly.com/question/26922844

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