A quadratic in vertex form reads as
(<em>x</em> - <em>a</em>)² + <em>b</em>
where (<em>a</em>, <em>b</em>) is the vertex.
To get the given quadratic in this form, complete the square:
<em>x</em>² - 6<em>x</em> - 40 = <em>x</em>² - 6<em>x</em> + 9 - 49 = (<em>x</em> - 3)² - 49
Or, work backwards by expanding the vertex form and solving for <em>a</em> and <em>b</em> :
(<em>x</em> - <em>a</em>)² + <em>b</em> = <em>x</em>² - 2<em>ax</em> + <em>a</em>² + <em>b</em>
So if
<em>x</em>² - 6<em>x</em> - 40 = <em>x</em>² - 2<em>ax</em> + <em>a</em>² + <em>b</em>,
then
-2<em>a</em> = -6 → <em>a</em> = 3
<em>a</em>² + <em>b</em> = -40 → <em>b</em> = -49
The value of x that will result for the rational expression to be undefined will be the values that will give 0 value for the denominator.
A. 7
<span>2(7)^2 -32 = 66
</span>
<span>B. 4
</span>2(4)^2 -32 = 0 <span>-------> undefined</span>
C. 2
2(2)^2 -32 = -24
<span>D. -2
</span>2(-2)^2 -32 = -24
<span>E. -4
</span>2(-4)^2 -32 = 0 -------> undefined
F. -7
2(-7)^2 -32 = 66
Answer:
D is the answer on edg.
Step-by-step explanation:
Answer:
true, false, true, true
Step-by-step explanation:
The set names in this diagram have nothing to do with exponents, radicals, and polynomials. We'll take the diagram at face value.
(a) The labels on the sets seem to be appropriately placed.
__
(b) "Some" in this context means "any or all of the set". Since all of the circle representing integers is outside the rectangle representing irrational numbers, it is TRUE that some integer are not irrational numbers.
No part of the circle representing whole numbers is inside the rectangle representing irrational numbers, so it is FALSE that some whole numbers are irrational numbers.
A portion of the circle representing integers is outside the circle representing whole numbers, so it is TRUE that some integers are not whole numbers.
Every part of the circle representing whole numbers is inside the rectangle representing rational numbers, so it is TRUE that all whole numbers are rational numbers.
