A wave with a frequency of 60 hertz would generate 60 wave crests every

What is the answer to -104=8x how do you find what x is

dybincka [34]

8x = -104 (I like to flip mine around like this, but it doesn't matter).
Now, you subtract 8 from both sides because you are doing the inverse.
-104 - 8 = -112
x=-112
Hope this helps!
Have a nice day!

7 0

3 years ago

Solve 0 = (x – 4)2 – 1 by graphing the related function.

fomenos

Answer:

Therefore, the solutions of the quadratic equations are:

$x=5,\:x=3$

The graph is also attached.

Step-by-step explanation:

The solution of the graph could be obtained by finding the x-intercept.

$y=\left(x-\:4\right)^2-1$

Finding the x-intercept by substituting the value y = 0

so

$y=\left(x-\:4\right)^2-1$

$\:0\:=\:\left(x\:-\:4\right)^2\:-\:1$ ∵ y = 0

$\left(x-4\right)^2-1=0$

$\left(x-4\right)^2-1+1=0+1$

$\left(x-4\right)^2=1$

$\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$\mathrm{Solve\:}\:x-4=\sqrt{1}$

$\mathrm{Apply\:rule}\:\sqrt{1}=1$

$x-4=1$

$x=5$

$\mathrm{Solve\:}\:x-4=-\sqrt{1}$

$\mathrm{Apply\:rule}\:\sqrt{1}=1$

$x-4=-1$

$x=3$

So, when y = 0, then x values are 3, and 5.

Therefore, the solutions of the quadratic equations are:

$x=5,\:x=3$

The graph is also attached. As the graph is a Parabola. It is visible from the graph that the values of y = 0 at x = 5 and x = 3. As the graph is a Parabola.

8 0

3 years ago

Help me with this question plz! I am really confused n strangled in it! although it seems easy I am not able to do...

SVEN [57.7K]

Answer:

please type out question as unclear to see :)

3 0

3 years ago

A professor has recorded exam grades for 10 students in his class, but one of the grades is no longer readable. If the mean sco

Nostrana [21]

Answer:

unreadable score = 35

Step-by-step explanation:

We are trying to find the score of one exam that is no longer readable, let's give that score the name "x". we can also give the addition of the rest of 9 readable s scores the letter "R".

There are two things we know, and for which we are going to create equations containing the unknowns "x", and "R":

1) The mean score of ALL exams (including the unreadable one) is 80

so the equation to represent this statement is:

mean of ALL exams = 80

By writing the mean of ALL scores (as the total of all scores added including "x") we can re-write the equation as:

$\frac{R+x}{10} =80$

since the mean is the addition of all values divided the total number of exams.

in a similar way we can write what the mean of just the readable exams is:

$\frac{R}{9} = 85\\$ (notice that this time we don't include the grade x in the addition, and we divide by 9 instead of 10 because only 9 exams are being considered for this mean.