2 years ago

Solve each system by elimination. (Help! I’ll mark you brainly)

Mathematics

1 answer:

zheka24 [161]2 years ago

4 0

Answer:

2) y=-1 4)y=-1

Step-by-step explanation

4x+8y=20

+ -4x+2y=-30

---------------------

10y=-10

isolate y by dividing 10 on each side

y=-1

-6x+5y=1

+ 6x+4y=-10

----------------------

9y=-9

y=-1

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2 years ago

Given: KL ║ NM , LM = 45, m∠M = 50° KN ⊥ NM , NL ⊥ LM Find: KN and KL

Mice21 [21]

Answer:

$KL=45\tan 50^{\circ}\sin 50^{\circ}\approx 41.08\\ \\KN=45\sin 50^{\circ}\approx 34.47$

Step-by-step explanation:

Given:

KL ║ NM ,

LM = 45

m∠M = 50°

KN ⊥ NM

NL ⊥ LM

Find: KN and KL

1. Consider triangle NLM. This is a right triangle, because NL ⊥ LM. In this triangle,

LM = 45

m∠M = 50°

So,

$\tan \angle M=\dfrac{\text{opposite leg}}{\text{adjacent leg}}=\dfrac{NL}{LM}=\dfrac{NL}{45}\\ \\NL=45\tan 50^{\circ}$

Also

$m\angle LNM=90^{\circ}-50^{\circ}=40^{\circ}$ (angles LNM and M are complementary).

2. Consider triangle NKL. This is a right triangle, because KN ⊥ NM . In this triangle,

$NL=45\tan 50^{\circ}$

$m\angle KLN=m\angle LNM=40^{\circ}$ (alternate interior angles)

$m\angle KNL=90^{\circ}-40^{\circ}=50^{\circ}$ (angles KNL and KLN are complementary).

So,

$\sin \angle KNL=\dfrac{\text{opposite leg}}{\text{hypotenuse}}=\dfrac{KL}{LN}=\dfrac{KL}{45\tan 50^{\circ}}\\ \\KL=45\tan 50^{\circ}\sin 50^{\circ}\approx 41.08$

and

$\cos \angle KNL=\dfrac{\text{adjacent leg}}{\text{hypotenuse}}=\dfrac{KN}{LN}=\dfrac{KN}{45\tan 50^{\circ}}\\ \\KN=45\tan 50^{\circ}\cos 50^{\circ}=45\sin 50^{\circ}\approx 34.47$