How can you use the Pythagorean theorem to solve problems

Mathematics

1 answer:

riadik2000 [5.3K]3 years ago

7 0

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adelina 88 [10]

Answer:

m= 6/-5

Step-by-step explanation:

m= y2-y1/x2-x1

m= 6-(-6)/-18-(-8)

m= 12/-10

m= 6/-5

8 0

3 years ago

). If a, ß are zeroes of the quadratic polynomial p(x)=kx²+4x+4 such

natta225 [31]

Answer:

The values of k are 2/3 and -1

Step-by-step explanation:

Product of zeros = αβ= constant / coefficient of x^2 = 4/k

Sum of zeros =α+β = - coefficient of x / coefficient of x^2= -4/k

Given

Consider a= α and b= β

$(\alpha)^2 + (\beta)^2 = 24$

$(\alpha)^2 + (\beta)^2$ can be written as $(\alpha)^2 + 2(\alpha)(\beta) + (\beta)^2$ if we add $\pm 2 (\alpha)(\beta)$ in the above equation.

$(\alpha)^2 + 2(\alpha)(\beta) + (\beta)^2 -2(\alpha)(\beta)$

$(\alpha + \beta)^2 -2(\alpha)(\beta)$

Putting values of αβ and α+β

$(\frac {-4}{k})^2 -2( \frac {4}{k}) = 24\\\frac {16}{k^2} - \frac {8}{k} = 24\\Multiplying\,\, the \,\, equation\,\, with\,\, 8K^2\\ 2 - k= 3K^2\\3k^2-2+k=0\\or\\3k^2+k-2=0\\3k^2+3k-2k-2=0\\3k(k+1)-2(k+1)=0\\(3k-2)(k+1)=0\\3k-2=0 \,\,and\,\, k+1 =0\\k= 2/3 \,\,and\,\, k=-1$