A set of 3 consecutive integers has a sum of 27. which integers are they

When you multiply a pair of conjugates you get a real number

marusya05 [52]

Answer: The complex conjugate has a very special property. Consider what happens when we multiply a complex number by its complex conjugate. We find that the answer is a purely real number it has no imaginary part. This always happens when a complex number is multiplied by its conjugate the result is real number.

Real numbers can be positive or negative, and include the number zero. They are called real numbers because they are not imaginary, which is a different system of numbers

Step-by-step explanation: Every complex number has a complex conjugate. The complex conjugate of a + bi is a - bi. For example, the conjugate of 3 + 15i is 3 - 15i, and the conjugate of 5 - 6i is 5 + 6i.

4 0

3 years ago

Trigonometry questions. PLEASE HELP WILL GIVE LOTS OF POINTS

nalin [4]

Step-by-step explanation:

It is all done I am not sure if it is correct

7 0

3 years ago

Why do you think Priya and Noah may have found different measurements for the same desk?

9966 [12]

Answer:

Priya used cm to measure the desk while Noah used inch to measure the desk that is why they found different measurement for same desk.

5 0

3 years ago

What is the best approximation of the projection of (5,-1) onto (2,6)?

Hatshy [7]

Answer:

Hence, the scalar projection of $\vec a$ onto $\vec b$ = $\frac{\sqrt{10} }{5}$ , and the vector projection of $\vec a$ onto $\vec b$ = $\frac{1}{5} \hat i+\frac{3}{5} \hat j$ .

Step-by-step explanation:

We have given two points $(5, -1)$ and $(2, 6)$ .

Let, $\vec a=5\hat {i}-\hat {j}$ and $\vec b= 2\hat {i}+6\hat{j}$ .

and we have calculate the projection of $\vec a$ onto $\vec b$ .

Now,

For the calculation of projection, first we need to calculate the dot product of $\vec a$ and $\vec b$ .

$\vec a.\vec b=(5\hat {i}-\hat{j}).(2\hat{i}+6\hat{j})$

$=10-6$

$=4$

then, we have to calculate the magnitude of $\vec b$ .

$\mid {\vec {b}}\mid$ = $\sqrt{2^{2}+6^{2} }$ = $\sqrt{40}$ = $2\sqrt{10}$ .

Now, the scalar projection of $\vec a$ onto $\vec b$ = $\frac{\vec a.\vec b}{\mid b\mid}$

= $\frac{4}{2\sqrt{10} }$ $\frac{2}{\sqrt{10} } \times\frac{\sqrt{10} }{\sqrt{10} } =\frac{2\sqrt{10} }{10} = \frac{\sqrt{10} }{5}$

and the vector projection of $\vec a$ onto $\vec b$ = $\frac{\vec a. \vec b}{\mid\vec b \mid^{2} } . \vec b$

= $\frac{4}{40} . (2\hat i+ 6\hat j)$

= $\frac{1}{5} \hat i+\frac{3}{5} \hat j$

Hence, the scalar projection of $\vec a$ onto $\vec b$ = $\frac{\sqrt{10} }{5}$ , and the vector projection of $\vec a$ onto $\vec b$ = $\frac{1}{5} \hat i+\frac{3}{5} \hat j$ .