On a common plane, two parallel lines do not intersect.
As a result, the characteristics of parallel lines with respect to transversals are given below.

For the give question;

Two line are cut by the transversal.

∠1 = 90.5° and ∠7 = 89.5°

Thus the result for the given statement are-

A. ∠4 is congruent to ∠5 because they are alternate interior angles; True.

B. Two lines are parallel; True.

C. The measure of ∠6 = 90.5°; False.

∠6 = ∠7 = 89.5°.(correct)

D. ∠2 and ∠3 are supplementary because they are same-side exterior Angeles; True.

Thus, the result for the given statement are found.

To know more about the parallel lines, here

#SPJ1

3 0

1 year ago

inysia [295]

Poetry is a type of literature, or artistic writing, that attempts to stir a reader’s imagination or emotions.

3 0

3 years ago

Mice21 [21]

Answer:

The given statement that value 5 is an upper bound for the zeros of the function f(x) = x⁴ + x³ - 11x² - 9x + 18 will be true.

Step-by-step explanation:

Given

$f\left(x\right)\:=\:x^2\:+\:x^3\:-\:11x^2\:-\:9x\:+\:18$

We know the rational zeros theorem such as:

if $x=c$ is a zero of the function $f(x)$ ,

then $f(c) = 0$ .

As the $f\left(x\right)\:=\:x^2\:+\:x^3\:-\:11x^2\:-\:9x\:+\:18$ is a polynomial of degree $4$ , hence it can not have more than $4$ real zeros.

Let us put certain values in the function,

$f(5) = 448$ , $f(4) = 126$ , $f(3) = 0$ , $f(2) = -20$ ,

$f(1) = 0$ , $f(0) = 18$ , $f(-1) = 16$ , $f(-2) = 0$ , $f(-3) = 0$

From the above calculation results, we determined that $4$ zeros as

$x = -3, -2, 1,$ and $3$ .

Hence, we can check that

$f(x) = (x+3)(x+2)(x-1)(x-3)$

Observe that,

$for x > 3$ , $f(x)$ increases rapidly, so there will be no zeros for $x>3$ .

Therefore, the given statement that value 5 is an upper bound for the zeros of the function f(x) = x⁴ + x³ - 11x² - 9x + 18 will be true.

5 0

3 years ago