The value of x<em> </em>in the polynomial fraction 3/((x-4)•(x-7)) + 6/((x-7)•(x-13)) + 15/((x-13)•(x-28)) - 1/(x-28) = -1/20 is <em>x </em>= 24
<h3>How can the polynomial with fractions be simplified to find<em> </em><em>x</em>?</h3>
The given equation is presented as follows;

Factoring the common denominator, we have;

Simplifying the numerator of the right hand side using a graphing calculator, we get;
By expanding and collecting, the terms of the numerator gives;
-(x³ - 48•x + 651•x - 2548)
Given that the terms of the numerator have several factors in common, we get;
-(x³ - 48•x + 651•x - 2548) = -(x-7)•(x-28)•(x-13)
Which gives;

Which gives;

x - 4 = 20
Therefore;
Learn more about polynomials with fractions here:
brainly.com/question/12262414
#SPJ1
The mapping rule for 90 degrees counterclockwise would be (-y,x). So if you had (x,y) it would transfer to (-y,x). Yours is wrong it should be (4,-3). Remember
-(-4) is 4 and -3 stays as -3.
Basically, memorize these mapping rules.
Hope this helped!
Hi,
Here we are going to be working on isolating the variable y, and seeing what its value equates to.
To do this, we must try and get the variable y on one side of the equation by itself.
Let's look at step one -
<em>4y - 1 = 7
</em>
We want to get rid of the 1 since we need to isolate x. We do this by doing the inverse of its operation. Since 1 is negative, if we add positive 1 to it - we will get 0, thereby being closer to isolating y.
However, when we do something on one side of the equation we must do it on the other. This means we will add 1 on both sides.
<em>4y - 1 + 1 = 7 + 1
</em>
<em>4y = 8
</em>
<em />Remember how I mentioned we do the inverse of the operation? In this case, 4 is multiplying y. The inverse operation of multiplication is division. So, to get rid of the 4 - we must divide 4y by 4, on both sides.
<em>4y / 4 = 8 / 4
</em>
<em>y = 2
</em>
We now know the variable y is equal to 2.
Hopefully, this helps.
Answer:
so the answer would be A. 7/5x +2
if you try graphing all the other equations it would be either over the picture or just not in the right place.