The frequency of the musical note b4 is about 493.88 hz.

what is the frequency of the note one octave above b4?

987.76 hz

740.82 hz

329.25 hz

246.9 hz

987.76 Hz

Step-by-step explanation:

You have B4's frequency of 493.88 Hz and you want to know the frequency of the note one octave above B4.

When you go up an octave, you double the original frequency.  When you go down an octave, you divide the original frequency by two.

So, since B4's frequency is about 493.88 Hz, an octave above would be 2 x 493.88 Hz or 987.76 Hz

If we were looking for an octave below B4, we would divide 493.88 Hz by 2 to get 246.94 Hz.

no, the equation would be invalid since they are both not equal.

step-by-step explanation:

here you got

4y-6+4y 2-9

first, you would simplify the equation on the form.

4y-6+4y 2-9

then, move all the terms that contains a y to the left and all the other terms to the right.

+4y-4y+-9-6

finally, you will simplify the left and the right side of the equation.

y=-3

the value of v.w= -12.

step-by-step explanation:

we are given three vector r,v and w as:

r = < 5, -5, -2> ; v = < 2, -8, -8> ; w = < -2, 6, -5>

we know that r,v and w could also be represented as:

r=5i-5j-2k

v=2i-8j-8k

and w= -2i+6j-5k

where i,j and k are the unit vectors of x,y and z-axis respectively.

also we know that it has the following property:

i.i=1 ; j.j=1; k.k=1

and i.=j.i=i.k=k.i=j.k=k.j=0

i.e. when we multiply two vectors or we find it's dot product we need to only multiply the components of x coordinates with each other; y with each other and z with each other.

hence, v.w= (2i-8j- -2i+6j-5k)

                  = 2×(-2)+(-8)×(6)+(-8)×(-5)

                  = -4-48+40

                  = -12

hence, v.w= -12