(ii) tan theta + sec theta - 1/tan theta - sec theta +1 =

1 + sin theta/cos theta

Skip to content# (ii) tan theta + sec theta – 1/tan theta – sec theta +1 = 1 + sin theta/cos theta

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(ii) tan theta + sec theta - 1/tan theta - sec theta +1 =

1 + sin theta/cos theta

answer: 33 ( approx)

step-by-step explanation:

since, in the linear regression formula, the slope is the a in the equation y = b + ax ( by comparing it with the general equation of line y = mx + c )

where, [tex]a = \frac{\sum y \sum x^2 - \sum x\sum xy}{n\sum x^2 - (\sum x)^2}[/tex]

here given table is,

price ($), x : 18 23 25 29 30 33

pairs sold, y : 16 13 10 8 6 2

so, [tex]\sum x = 158[/tex] , [tex]\sum y = 55[/tex], [tex]\sum xy = 1315[/tex],

[tex]\sum x^2 = 4308[/tex] and [tex]\sum y^2 = 629[/tex],

thus, slop of the given line is,

[tex]a = \frac{55\times 4308-158\times 1315}{6\times 4308-(158)^2}[/tex]

⇒ [tex]a = \frac{29170}{884}[/tex]

⇒ [tex]a =32.9977375566[/tex]≈ 33

the simplified form is

[tex]\frac{x^2(2x+3)}{2(4x-3)}[/tex]

step-by-step explanation:

we want to simplify the expression,

[tex]\frac{\frac{6x+9}{15x^2} }{\frac{16x-12}{10x^4} }[/tex]

let us change the middle bar to a normal division sign to obtain,

[tex]\frac{6x+9}{15x^2}\div\frac{16x-12}{10x^4}[/tex]

we multiply the first fraction by the reciprocal of the second fraction to obtain,

[tex]\frac{6x+9}{15x^2}\times \frac{10x^4}{16x-12}[/tex]

we factor to obtain,

[tex]\frac{3(2x+3)}{3\times5\times x^2}\times \frac{2\times5\times x^2\times x^2}{4(4x-3)}[/tex]

we cancel the common factors to get,

[tex]\frac{(2x+3)}{1}\times \frac{ x^2}{2(4x-3)}[/tex]

we simplify to get,

[tex]\frac{x^2(2x+3)}{2(4x-3)}[/tex]

the correct answer is b

The answer would be a

step-by-step explanation:

1. the cylinder is a. 54

2. a.108 /