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2 years ago

Super easy question HELP

Mathematics

1 answer:

Elena L [17]2 years ago

4 0

It's A. 0.5
0.5*0.5*0.5= 0.125
cube root the volume.

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Answer:

in geometry nation, is circle, isosceles triangle and parabola.

Step-by-step explanation:

just did it.

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2 years ago

Rachel purchased a prepaid phone card for $30. Long distance calls cost 6 cents a minute using this card. Rachel used her card o

Amiraneli [1.4K]

Answer: her call lasted for 26 minutes.

Step-by-step explanation:

Let x represent the number of minutes for which her call lasted.

Rachel purchased a prepaid phone card for $30. Long distance calls cost 6 cents a minute using this card. Converting 6 cents to dollars, it becomes 6/100 = $0.06

This means that the cost of x minutes of long distance call is

0.06 × x = $0.06x.

If the remaining credit on her card is $28.44, it means that

0.06x + 28.44 = 30

0.06x = 30 - 28.44

0.06x = 1.56

x = 1.56/0.06

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2 years ago

The first step in solving 7/8 = 3x - 5/11 is to subtract 5/11 from both sides of the equation.

Ostrovityanka [42]

False because -5/11 should be add on both sides

3 0

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

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

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Answer:

the first and the second answer

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