Select all expressions that are equivalent to − 4 ( x + 2 ) − 2 x + 4 .

Mathematics

1 answer:

salantis [7]3 years ago

4 0

Answer:

A), D), E), F).

Step-by-step explanation:

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nevsk [136]

Answer:

46 cm

Step-by-step explanation:

irsitsitdoyft8soyffyoyrafsiigzirsdifogstistsi

4 0

3 years ago

Factorising quadratics X squared + 5x -6 X squared - 2x15 X squared -7x +12

34kurt

Answer:

(x -1)(x +6)
(x -5)(x +3)
(x -4)(x -3)

Step-by-step explanation:

In each case, you're looking for divisors of the constant that have a sum equal to the x-coefficient. Those divisors are the constants in the binomial factors.

1) -6 = -1·6 = -2·3 . . . . . (-1)+(6) = 5, so these are the constants of interest.

x^2 +5x -6 = (x -1)(x +6)

2) -15 = -1·15 = -3·5 = -5·3 . . . . . (-5)+(3) = -2, so these are the constants of interest

x^2 -2x -15 = (x -5)(x +3)

3) 12 = -1·-12 = -2·-6 = -3·-4 . . . . . (-3) +(-4) = -7, so these are the constants of interest

x^2 -7x +12 = (x -3)(x -4)

5 0

3 years ago

Write an expression for the 12th partial sum of the series 3/2+7/3+19/6+... using summation notation

lapo4ka [179]

Answer:

$S_{12}=\sum_{i=1}^{12} [\frac{3}{2}+(i-1)\times \frac{5}{6}]$

$S_{12}=73$

Step-by-step explanation:

$First\ term\ of\ the\ series(a_1)=\frac{3}{2}\\\\Second\ term\ of\ the\ series(a_2)=\frac{7}{3}\\\\Third\ term\ of\ the\ series(a_3)=\frac{19}{6}\\\\a_2-a_1=\frac{7}{3}-\frac{3}{2}=\frac{5}{6}\\\\a_3-a_2=\frac{19}{6}-\frac{7}{3}=\frac{5}{6}\\\\Hence\ it\ is\ an\ Arithmetic\ Series\ with\ first\ term=\frac{3}{2}\ and\ constant\ difference=\frac{5}{6}$

$a_1=\frac{3}{2}+0\times \frac{5}{6}\\\\a_2=\frac{3}{2}+1\times \frac{5}{6}\\\\a_3=\frac{3}{2}+2\times \frac{5}{6}\\\\.\\.\\.\\a_n=\frac{3}{2}+(n-1)\times \frac{5}{6}\\\\S_n=a_1+a_2+a_3+......+a_n\\\\S_n=(\frac{3}{2}+0\times \frac{5}{6})+(\frac{3}{2}+1\times \frac{5}{6})+(\frac{3}{2}+2\times \frac{5}{6})+....+(\frac{3}{2}+[n-1]\times \frac{5}{6})\\\\S_n=\sum_{i=1}^n [\frac{3}{2}+(i-1)\times \frac{5}{6}]\\\\S_n=(\frac{3}{2}+\frac{3}{2}+\frac{3}{2}+...n\ times)+\frac{5}{6}(1+2+3+4+...+(n-1))\\\\$