answer:Double spotting in a receiver occurs when two closely spaced signals are received simultaneously, creating an overlapping or double image on the display. This can be primarily attributed to: C) Inadequate image frequency rejection, which allows unwanted signals from the image frequency to interfere with the desired signal. While other factors might influence the receiver's performance, such as sensitivity of the RF amplifier and gain of the IF amplifier, the specific issue of double spotting is most closely related to the receiver's ability to reject image frequencies.

answer:The divergence of vec{F} is given by text{div} vec{F} = frac{partial}{partial x} f(x,y,z) + frac{partial}{partial y} g(x,y,z) + frac{partial}{partial z} h(x,y,z) = frac{partial}{partial x} ln (y-x) + frac{partial}{partial y} sqrt{y} + frac{partial}{partial z} tan (z) = -frac{1}{y-x} + frac{1}{2sqrt{y}} + sec^2 (z) Therefore, the divergence of vec{F} is -frac{1}{y-x} + frac{1}{2sqrt{y}} + sec^2 (z).

answer:The equation of a line in #color(blue)"slope-intercept form"# is: #color(red)(bar(ul(|color(white)(2/2)color(black)(y=mx+c)color(white)(2/2)|)))# where m is the slope and c is the y-intercept. #"Rearrange " 2x-5y+10=0 " into this form"# #"subtract 2x from both sides"# #-5y+10=-2x# #"subtract 10 from both sides"# #-5y=-2x-10# #"divide both sides by -5"# #y=-2/-5x-10/-5# #rArry=2/5x+2larrcolor(red)" in slope-intercept form"#

answer:Given vertices of triangle are P(1, 2, 3), Q(-2, 1, 4), and R(0, 3, 2). First, we find two new vectors using the given points: begin{align*} overrightarrow{PQ} &= overrightarrow{Q} - overrightarrow{P} &= left langle -2, 1, 4 right rangle - left langle 1, 2, 3 right rangle &= left langle -3, -1, 1 right rangle overrightarrow{PR} &= overrightarrow{R} - overrightarrow{P} &= left langle 0, 3, 2 right rangle - left langle 1, 2, 3 right rangle &= left langle -1, 1, -1 right rangle end{align*} Now, we find the vector product of two vectors: begin{align*} overrightarrow{PQ} times overrightarrow{PR} &= begin{vmatrix} vec{i} & vec{j} & vec{k} -3 & -1 & 1 -1 & 1 & -1 end{vmatrix} &= (1 - 1) vec{i} - (-3 - 1) vec{j} + (-3 + 1) vec{k} &= 4 vec{j} - 2 vec{k} end{align*} Finally, we find the area of the triangle: begin{align*} text{Area} &= frac{1}{2} left | overrightarrow{PQ} times overrightarrow{PR} right | &= frac{1}{2} sqrt{(0)^2 + (4)^2 + (-2)^2} &= frac{1}{2} sqrt{20} &= boxed{sqrt{5}} end{align*} Therefore, the area of the triangle is sqrt{5}.