Explain which theorems, definitions, or combinations of both can be used to prove that alternate exterior angles are congruent.

Answers

Answer 1

Answer: 1. The first theorem used is that vertical angles are congruent.
2. The next theorem used is that adjacent angles in a parallelogram are supplementary.
3. The definition of supplementary angles is then used for angle formed by intersecting lines.
4. The theorem on vertical angles is used again.
5. Finally, the definition of the transitivity property is used to prove that alternate exterior angles are congruent.

Answer 2

Answer:

Using the Corresponding Angles Theorem, Vertical Angles Theorem, and the Transitive Property of Congruence, we can prove that alternate exterior angles (e.g, <4 and <5) are congruent by the alternate exterior angles theorem.

Recall:

Alternate exterior angles are angles that lie outside the two lines that is cut across by a transversal but on opposite sides along the transversal.
Examples of alternate exterior angles are <2 and <7; <4 and <5 as shown in the figure attached below.

If we are given that $m \parallel n$ in the diagram attached below, the following are theorems and definitions we can use to prove that $\angle 4 \cong \angle 5$ (alternate exterior angles).

Statement 1: $\angle 4 \cong \angle 8$

Reason: Corresponding Angles Theorem

The corresponding angles theorem states that when two parallel lines (lines m and n) are intersected by a transversal line (line w), the two corresponding angles formed (e.g. <4 and <8) are congruent.

Statement 2: $\angle 8 \cong \angle 5$

Reason: Vertical Angles Theorem

The Vertical Angles Theorem states that the opposite vertical angles (e.g. <8 and <5) formed when two lines (lines n and w) intersect are congruent to each other.

Statement 3: $\angle 4 \cong \angle 5$

Reason: Transitive Property of Congruence

The Transitive Property of Congruence states that if a = b; and b = c; then a = c.

Therefore, using the Corresponding Angles Theorem, Vertical Angles Theorem, and the Transitive Property of Congruence, we can prove that alternate exterior angles (e.g, <4 and <5) are congruent by the alternate exterior angles theorem.

Learn more here:

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Rewrite the expression x^3+10x^2+13x+39/x^2+2x+1 in the form of q(x)+r(x)/b(x)

Answers

$(x^3+10x^2+13x+39)/(x^2+2x+1)$

$x^3=x\cdot x^2$ , and $x(x^2+2x+1)=x^3+2x^2+x$ . Subtracting this from the numerator gives a remainder of

$(x^3+10x^2+13x+39)-(x^3+2x^2+x)=8x^2+12x+39$

$8x^2=8\cdot x^2$ , and $8(x^2+2x+1)=8x^2+16x+8$ . Subtracting this from the previous remainder gives a new remainder of

$(8x^2+12x+39)-(8x^2+16x+8)=-4x+31$

$-84x$ is not a multiple of $x^2$ , so we're done. Then

$(x^3+10x^2+13x+39)/(x^2+2x+1)=x+8+(-4x+31)/(x^2+2x+1)$

(a) Solve 7( k - 3 ) = 3k - 5 (b) Expand and simplify (2x + 3 )( x - 8)

(c) Solve 7 - 3 f = 2
4

Answers

a) $7(k-3)=3k-5\n 7k-21=3k-5\n 7k-3k=-5+21\n 4k=16\n k=\frac { 16 }{ 4 } \n k=4$

b) $(2x+3)(x-8)\n 2{ x }^( 2 )-16x+3x-24\n 2{ x }^( 2 )-13x-24$

c) $\frac { 7-3f }{ 4 } =2\n 7-3f=4\cdot 2\n 7-3f=8\n -3f=8-7\n -3f=1\n f=-\frac { 1 }{ 3 }$

$(a)\n7(k-3)=3k-5\n7(k)+7(-3)=3k-5\n7k-21=3k-5\ \ \ \ |add\ 21\ to\ both\ sides\n7k=3k+16\ \ \ \ |subtract\ 3k\ from\ both\ sides\n4k=16\ \ \ \ \ |divide\ both\ sides\ by\ 4\n\boxed{k=4}$

$(b)\n(2x+3)(x-8)=(2x)(x)+(2x)(-8)+(3)(x)+3(-8)\n\n=2x^2-16x+3x-24=\boxed{2x^2-13x-24}$

$(c)\n(7-3f)/(4)=2\ \ \ \ |multiply\ both\ sides\ by\ 4\n\n\not4^1\cdot(7-3f)/(\not4_1)=4\cdot2\n\n7-3f=8\ \ \ \ \ |subtract\ 7\ from\ both\ sides\n\n-3f=1\ \ \ \ \ |divide\ both\ sides\ by\ (-3)\n\n\boxed{f=-(1)/(3)}$

B and C? Please? I already did the table

Answers

Now you're trying hard to make something hard out of something easy.

It gives you the equation . . . C = -7.11 t + 153.9

b). Take this equation, and calculate 'C' when 't' is -20.

   Find 'C' : C = (-7.11)(-20) + 153.9

   I'm sure you can do this easily.

c). Take this equation, and calculate what 't' would have to be
   in order to have C=50 .

   Find 't': 50 = (-7.11) t + 153.9

       I'm sure you can do this easily too.

From a fraction to a decimal 19/20