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<h2>______________________________</h2><h2>Solve for x | Solution and Explanation </h2>
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Hello! So...
We are given the following:
Solve for x.
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1. Group the like terms.
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2. Add similar elements ().
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3. Subtract 23 from both sides.
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4. Simplify.
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5. Divide both sides by 4.
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6. Simplify.
(aka. Option A)
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Hope this helps!
Answer: An integer added to an integer is an integer, this statement is always true. A polynomial subtracted from a polynomial is a polynomial, this statement is always true. A polynomial divided by a polynomial is a polynomial, this statement is sometimes true. A polynomial multiplied by a polynomial is a polynomial, this statement is always true.
Explanation:
1)
The closure property of integer states that the addition, subtraction and multiplication is integers is always an integer.
If , then a+b\in Z.
Therefore, an integer added to an integer is an integer, this statement is always true.
2)
A polynomial is in the form of,
Where are constant coefficient.
When we subtract the two polynomial then the resultant is also a polynomial form.
Therefore, a polynomial subtracted from a polynomial is a polynomial, this statement is always true.
3)
If a polynomial divided by a polynomial then it may or may not be a polynomial.
If the degree of numerator polynomial is higher than the degree of denominator polynomial then it may be a polynomial.
For example:
Then , which a polynomial.
If the degree of numerator polynomial is less than the degree of denominator polynomial then it is a rational function.
For example:
Then , which a not a polynomial.
Therefore, a polynomial divided by a polynomial is a polynomial, this statement is sometimes true.
4)
As we know a polynomial is in the form of,
Where are constant coefficient.
When we multiply the two polynomial, the degree of the resultand function is addition of degree of both polyminals and the resultant is also a polynomial form.
Therefore, a polynomial subtracted from a polynomial is a polynomial, this statement is always true.
Point D is the midpoint of the outer circle that we aim to find the area of
The circle has a diameter of WZ and radii of WC and CZ
We know that YZ=YD=10 cm
Let DC be
and CY be 
The radius of the outer circle can be written as
or 
which we can equate to find the value of
Therefore, the radius of the circle is
And hence the area of the circle is
=324
The circle has a diameter of WZ and radii of WC and CZ
We know that YZ=YD=10 cm
Let DC be
The radius of the outer circle can be written as
Therefore, the radius of the circle is
And hence the area of the circle is