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3 years ago

Plsss helppp asappp will give brainliest

Mathematics

1 answer:

hichkok12 [17]3 years ago

4 0

Just multiply 7.20 x 5 since the giving N U T S Are 5 15 pounds nuts for 7.20

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Please help me!! Also if you can show the steps, I’d appreciate it. Thank you very much

natita [175]

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3 years ago

A photocopier costs $25 000. Its unit cost depreciation is calculated at $0.01 per page. The equation to determine its value aft

Drupady [299]

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3 years ago

Help please, I need help understanding this. I'm in Geometry

Rasek [7]

Answer:

I believe you can set the equations equal to each other and solve for x.

Step-by-step explanation:

reason being is that by definition, a line can be infinite, so in theory, those 2 lines can be equal.

8 0

3 years ago

Chad sells custom T-shirts with five of his friends. If the earnings are divided evenly, how much must the business make in orde

Ede4ka [16]

Answer:

I feel like the answer is B because they each want 45$ and they want it to be equal.

Step-by-step explanation:

I had the test

4 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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