Bolts are usually classified in terms of two types of measurements: the metric system and the imperial system. In calipers, the metric system is in units of millimeters, while the imperial system is in units of inches. For example, when a bolt is sized as 12 × 1.25 in the metric system, that means the bolt thread diameter measures 12 mm, and the thread pitch measures 1.25 mm.

So, if your metric bolt measures 0.78 over the threads, that means that the size of the bolt is 0.78 mm in thread pitch. The attached picture shows how you measure the thread pitch.

5 0

2 years ago

In the diagram below, line AB and line CD intersect at point P. Name the angle that is always congruent to <APC

Alex Ar [27]

<dpb because they’re vertical angles (they’re directly across from each other)

5 0

3 years ago

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Minus A plus 6 a ncert solution class 9

lianna [129]

Answer:

yes htjjevdhfifnfjjdeirjfjebdv dbdu you fr

5 0

2 years ago

What eigen value for this matix <br> (1 -2)<br> (-2 0)

natali 33 [55]

You find the eigenvalues of a matrix A by following these steps:

Compute the matrix $A' = A-\lambda I$ , where I is the identity matrix (1s on the diagonal, 0s elsewhere)
Compute the determinant of A'
Set the determinant of A' equal to zero and solve for lambda.

So, in this case, we have

$A = \left[\begin{array}{cc}1&-2\\-2&0\end{array}\right] \implies A'=\left[\begin{array}{cc}1&-2\\-2&0\end{array}\right]-\left[\begin{array}{cc}\lambda&0\\0&\lambda\end{array}\right]=\left[\begin{array}{cc}1-\lambda&-2\\-2&-\lambda\end{array}\right]$

The determinant of this matrix is

$\left|\begin{array}{cc}1-\lambda&-2\\-2&-\lambda\end{array}\right| = -\lambda(1-\lambda)-(-2)(-2) = \lambda^2-\lambda-4$

Finally, we have

$\lambda^2-\lambda-4=0 \iff \lambda = \dfrac{1\pm\sqrt{17}}{2}$

So, the two eigenvalues are

$\lambda_1 = \dfrac{1+\sqrt{17}}{2},\quad \lambda_2 = \dfrac{1-\sqrt{17}}{2}$

5 0

2 years ago

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