Simplify the following:
2^3 (4×36/12 + 1/4)

4×36/12 = (4×36)/12:
2^3 ((4×36)/12 + 1/4)
2^3 = 2×2^2:
2×2^2 ((4×36)/12 + 1/4)
2^2 = 4:
2×4 ((4×36)/12 + 1/4)
2×4 = 8:
8 ((4×36)/12 + 1/4)
36/12 = (12×3)/12 = 3:
8 (4×3 + 1/4)
4×3 = 12:
8 (12 + 1/4)
Put 1/4 + 12 over the common denominator 4. 1/4 + 12 = 1/4 + (4×12)/4:
8 1/4 + (4×12)/4
4×12 = 48:
8 (1/4 + 48/4)
1/4 + 48/4 = (1 + 48)/4:
8 (1 + 48)/4
1 + 48 = 49:
8×49/4

8×49/4 = (8×49)/4:
(8×49)/4
8/4 = (4×2)/4 = 2:
2×49
2×49 = 98:
Answer: 98

4 0

4 years ago

Solve for n , S = (n - 2)180

pogonyaev

S=(n-2)180
Dived 180 both sides
S/180=n-2 add 2 both sides
S/180+2=n
S over 182=n
Hope this helps good luck have a nice nite

7 0

3 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

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

3 0

3 years ago

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