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b. Turns per inch

The amount or degree of yarn twist is measured in number of TPI. full form of TPI is Turns per inch.

Correct answer is d) Worm and Nut

A. More than the glomerular filtration rate

As we know, time taken is inversely proportional to the speed if the distance covered is constant.

Thus, the ratio of the time taken by Madhur, Mukesh and Mitali = (1/7):(1/8):(1/9) = 72:63:56

Let time taken by Madhur, Mukesh and Mitali be 72x, 63x and 56x min respectively.

So, 72x + 63x + 56x = 95.5 ⇒ x = 95.5/191 = 0.5

Hence, the time taken by Mitali = 56x = 56 × 0.5 = 28 min

In atmosphere, oxygen is found as
Diatomic molecular form with chemical formula O₂ known as oxygen.
Triatomic molecular form with chemical formula O₃ known as ozone.
The presence of oxygen in the air (atmosphere) is very essential for the following:
1. The life processes such as respiration, in all living organisms.
2. Combustion of fuels. Otherwise, fire would not have been possible
The presence of ozone in the air (atmosphere) is very essential for the following:
1. Ozone is a gas in the atmosphere that protects everything living on the Earth from harmful ultraviolet (UV) rays from the Sun.

Given:

The ratio of gold and silver in an alloy = 7 ∶ 3 

The weight of the gold in the alloy = 61.6 gm

Calculations:

The weight of silver in the alloy = (61.6/7) × 3 = 8.8 × 3 = 26.4 gm

∴ The weight of silver in the alloy is 26.4 gm.

Difference between x coordinates of A and B = 3 – 2 = 1, 

Coordinate of y = 4 + 1 = 5 

x coordinates of D = –2 – 1 = -3

y coordinates of D = 3 – 5 = –2 

Coordinates of D = (–3, –2)

Equation of the circle

(x – 2)² + (x – 1)² = 5

x₂ – 4x + 4 + y₂  – 2y + 1 = 5

x₂ + y₂ – 4x – 2y = 0

As (0, 1), (2, 3) are the points joining the diameter of the circle, the center will be

\(=(\frac{0+2}{2},\frac{1+3}{2})=(1,2)\)

Radius = \(\sqrt{(1-0)^2+(2-1)^2}\)

\(\Rightarrow\) r = \(\sqrt{1^2+1^2}=\sqrt{2}\)

(x, y) lies on the cirlce

(x - 1)² + (y - 2)² = \(\sqrt{2}^2\)

x² - 2x + 1 + y² - 4y + 4 = 2

x² + y² - 2x - 4y + 5 - 2 = 0

x² + y² - 2x - 4y + 3 = 0

Let (x, 0) be the coordinate which intersect the x axis then, x² – 2x + 3 = 0

\(x=\frac{2\pm\sqrt{4-12}}{2}=\frac{2\pm\sqrt{-8}}{2}\)

This circle can not be intersect the x axis. 

Let (0, y) be the coordinate which intersect the y axis then, y – 4y + 3 = 0, (y – 3)(y – 1) = 0 

Coordinates are (0, 1) and (0, 3)

If (x, y) lies' on the line joining the points (0,3) and (3,0) then,

\(\frac{y-3}{x-0}=\frac{0-3}{3-0}\)

\(\Rightarrow \frac{y-3}{x}=\frac{-3}{3}=-1\)

y - 3 = x

\(\Rightarrow\) x + y = 3

∴ the sum of the and y coordinates of the point is 3.

One end of the diameter = (x , y ) = (1, 4) 

Center of the circle = (x, y) =(3, -4) If other end of the diameter = (x₂, y₂)

\(3=\frac{1+x_2}{2},\)

∴ x₂ = 6 - 1

= 5 - 4 = \(\frac{4+y_2}{2},\)

∴ y₂ = -8 - 4 = -12,

Other end of the diameter (5 – 12)

The distance between the centre and the point (-8,-6) is = \(\sqrt{(-8)^2+(-6)^2}\)

= \(\sqrt{100}\) = 10

It is equal to the radius. So the point (-8, -6) is a point on the circle.

The distance between the centre, and the point (9,-1) is = \(\sqrt{9^2+(-1)^2}\)

= \(\sqrt{81+1}\) = \(\sqrt{82}\)

√82 is smaller than 10.

So the point (9, –1) is a point inside the circle.

Distance between (x₁, y₁) and (x₃, y₂) is

\(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

Distance of (x, y) from (0, 0) is \(\sqrt{x^2+y^2}\)

The distance to the points (–2,–3) is \(\sqrt{(-2)^2+(-3)^2}=\sqrt{13}\)

The distance to the points (2, 3) is \(\sqrt{2^2+3^2}=\sqrt{13}\)

Both the distances are equal.

Equation of the circle

= (x – 0)² + (y – 0)² = 5²

= x² + y² = 25

x² = 25 – y²

x = \(\sqrt{25-y^2}\)

y = 0 x = 5 (5, 0), (–5, 0)

y = 5 x = 0 (0, 5), (0, –5)

y = 3 x = 4 (4, 3), (4, 3)

y = 4 x = 3 (3, 4), (3, –4)

Coordinates of points on the circle (5, 0), (–5, 0), (0, 5), (0, -5), (4, 3), (4, 3), (3, 4), (3, 4).

Points are (4, 0) (3, –5)

Slope = \(\frac{-5}{-1}=5\)

This line is parallel to the x-axis so the slope is 0.

Slope = \(\frac{-5+3}{3-1}=\frac{-2}{2}=1\)

Slope of the line = tan 60° = √3

Distance between (2, 7) and (5, 11) = \(\sqrt{9+16}=5\)

Distance between (5, 11) and (8, 15) = \(\sqrt{9+16}=5\)

Hence the distances are same, so the points are lie on the same line.

Slope of the line = \(\frac{-3-0}{0-5}=\frac{-3}{-5}=\frac{3}{5}\)

Slope of the line which passing through the points (–2, 3) and (5, 7) = \(\frac{7-3}{5+2}=\frac{4}{7}\)

The line which joins the points (–2, 2) and (5, 6) is parallel to the first line.

Slope of the line which passes through the points (–2, 2) and (5, 6) = \(\frac{6-2}{5+2}=\frac{4}{7}\)

Slope of the line which passes through the points (–7, 4) and (0, 8) = \(\frac{8-4}{0+7}=\frac{4}{7}\)

The lines are parallel because their slopes are same.

 \([1\,\,\,{-1}\,\,\,\,x]\left[\begin{matrix}0&1&-1\\2&1&3\\1&1&1\end{matrix}\right]\, \left[\begin{matrix}0\\1\\1\end{matrix}\right] =0\) 

⇒ \([1\,\,\,{-1}\,\,\,\,x]\left[\begin{matrix}0\\4\\2\end{matrix}\right] =0\)

⇒ - 4 + 2x = 0

⇒ x = \(\frac42 = 2\)

A = \(\begin{bmatrix}2&4\\3&2\end{bmatrix}\), B = \(\begin{bmatrix}-2&5\\3&4\end{bmatrix}\)

\(\therefore\) 3A - B = 3\(\begin{bmatrix}2&4\\3&2\end{bmatrix}\) - \(\begin{bmatrix}-2&5\\3&4\end{bmatrix}\)

 = \(\begin{bmatrix}6&12\\9&6\end{bmatrix}\) - \(\begin{bmatrix}-2&5\\3&4\end{bmatrix}\)

 = \(\begin{bmatrix}6-(-2)&12-5\\9-3&6-4\end{bmatrix}\) = \(\begin{bmatrix}8&7\\6&2\end{bmatrix}\)

Let A = \(\begin{bmatrix}1&3&-2\\-3&0&1\\2&1&0\end{bmatrix}\)

We have A = I A

⇒ \(\begin{bmatrix}1&3&-2\\-3&0&1\\2&1&0\end{bmatrix}\) = \(\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}A\) 

Applying R₂ → R₂ + 3R₁

R₃ → R₃ - 2R₁

\(\begin{bmatrix}1&3&-2\\0&9&-5\\0&-5&4\end{bmatrix}\)= \(\begin{bmatrix}1&0&0\\3&1&0\\-2&0&1\end{bmatrix}A\) 

Applying R₃ → 9R₃ + 5R₂

\(\begin{bmatrix}1&3&-2\\0&9&-5\\0&0&11\end{bmatrix}\) = \(\begin{bmatrix}1&0&0\\3&1&0\\-3&5&9\end{bmatrix}A\)

Applying R₃ → R₃/11

\(\begin{bmatrix}1&3&-2\\0&9&-5\\0&0&1\end{bmatrix}\) = \(\begin{bmatrix}1&0&0\\3&1&0\\-3/11&5/11&9/11\end{bmatrix}A\) 

Applying R₁ → R₁ + 2R₃

R₂  → R₂ + 5 R₃

\(\begin{bmatrix}1&3&0\\0&9&0\\0&0&1\end{bmatrix}\) = \(\begin{bmatrix}5/11&10/11&18/11\\18/11&36/11&45/11\\-3/11&5/11&9/11\end{bmatrix}A\) 

Applying R₂ → R₂/9

\(\begin{bmatrix}1&3&0\\0&1&0\\0&0&1\end{bmatrix}\) = \(\begin{bmatrix}5/11&10/11&18/11\\2/11&4/11&5/11\\-3/11&5/11&9/11\end{bmatrix}A\)

Applying R₁ → R₁ - 3R₂

\(\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\) = \(\begin{bmatrix}-1/11&-2/11&3/11\\2/11&4/11&5/11\\-3/11&5/11&9/11\end{bmatrix}A\)

We obtain I = \(\begin{bmatrix}-1/11&-2/11&3/11\\2/11&4/11&5/11\\-3/11&5/11&9/11\end{bmatrix}A\)

\(\therefore\) A^-1 = \(\begin{bmatrix}-1/11&-2/11&3/11\\2/11&4/11&5/11\\-3/11&5/11&9/11\end{bmatrix}\)(\(\because\) A^-1 A = I)

Scanning exceptions are lists of trusted or untrusted sites (hostnames and URLs) that are never analyzed or always analyzed. The type of analysis to never or always perform is specified per hostname or URL, or group of hostnames and URLs.

The correct option is  (d) CH3^- will go for pyramidal geometry as four electron pairs are distributed in a tetrahedral shape. If there is one lone pair of electrons and three bond pairs the resulting molecular geometry is trigonal pyramidal.

The close packed direction of a BCC crystal is  〈1 1 1〉