Mathematics, 06.04.2020 08:37 jj1077348

That
7. In the diagram, LMN = ONM= 90°. P is the midpoint of MN,
MN = 2ML and MN = NO. Prove that:
a) the triangles MNL and NOP are congruent
b) OÊN = LÑO
c) LỘO=90°

Answers: 1

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Answer from: rafaellazaro60pcw3w5

In the diagram,LMN = ONM= 90°. P is the midpoint of MN, MN = 2ML and

MN = NO

Step-by-step explanation:

Step 1

(A) Angle LMN = angle ONM [given]

MN = NO [given]

P is the midpoint of MN [given]

MN = 2 PN [definition of midpoint]

MN = 2 ML [given]

2 ML = 2 PN [transitivity of equality]

ML = PN [division property of equality]

Triangles MNL and NOP are congruent [side-angle-side]

Step 2

(B) Angles LMN and ONM measure 90° [given]

Measures of angles LMN, MNL, and NLM sum to 180° [angles of a triangle]

Measures of angles MNL and NLM sum to 90° [subtraction property of equality]

Measures of angles MNL and LNO sum to 90° [adjacent angles]

Measures of angles MNL and NLM = measures of angles MNL and LNO [transitive property of equality]

Measures of angles NLM and LNO are equal [subtraction property of equality]

Angles NLM and LNO are congruent [definition of congruent angles]

Triangles MNL and NOP are congruent [proved in part (a)]

Angles OPN and NLM are congruent [corresponding angles of congruent triangles]

Angles OPN and LNO are congruent [transitive property of angle congruence]

Angles NOQ and NOP are congruent [same angle]

Angles OPN and LNO are congruent [proved in part (b)]

Angles QNO and OPN are congruent [transitive property of angle congruence]

Triangles NOP and QON are similar [angle-angle similarity]

Angles NQO and PNO have the same measure [corresponding angles of similar triangles]

Angle MNO measures 90° [given]

Angle PNO measures 90° [same angle]

Angle NQO measures 90° [equal angles]

LN is perpendicular to PO [definition of perpendicular]

Angle LQO measures 90° [definition of perpendicular]

Thus we can say that LQO=90°

Answer from: Quest

x= -3 and $x=\frac{-3+/-3i\sqrt{3} }{2}$

step-by-step explanation:

to find the roots of a cubic, use division or factoring to find the factors then set equal to 0. as this is a difference of cubes, it factors into

$(x^3-a^3)=$ $(x-a)(x^2+ax+a^2)$

where a is the cube root of $a^3$ .

here this is 27 and the cube root of 27 is 3. so we write using the form:

$(x-3)(x^2+(3)x+3^2)\\(x-3)(x^2+3x+9)$ .

set each factor to 0 and solve for x.

x-3=0

x=3

the quadratic expression does not factor and must be solved using the quadratic formula.

$x=\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

where a = 1, b=3, and c=9

$x=\frac{-3+/-\sqrt{3^2-4(1)(9)} }{2(1)} \\x=\frac{-3+/-\sqrt{9-36)} }{2} \\x=\frac{-3+/-\sqrt{-27)} }{2}\\x=\frac{-3+/-3i\sqrt{3} }{2}$

this cubic has two complex roots.

Answer from: Quest

x=half

step-by-step explanation:

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7. In the diagram, LMN = ONM= 90°. P is the midpoint of MN,
MN = 2ML and MN = NO. P...

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