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ANEK [815]
3 years ago
5

PLEASE HELP ME!!!! PICTURE IS PROVIDED

Mathematics
1 answer:
sattari [20]3 years ago
6 0

Answer:

pretty sure C

Step-by-step explanation:

sorry if im incorrect information

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Change 3/2 to a mixed number
Mariulka [41]
Okay first you have to divide 
2 goes into 3
3/2
2 fits into 3 once
1 and the left over is 1 so
1 1/2

4 0
3 years ago
Which inequality would be in the correct slope-intercept form:
WINSTONCH [101]

Answer:

J. y > 3x + 5

Step-by-step explanation:

To convert the standard form equation into slope-intercept, we must make y isolated.

Step 1: Subtract 6x from both sides.

  • 6x-2y-6x
  • -2y

Step 2: Divide both sides by -2 and flip the inequality sign.

  • -2y/-2 < (-6x-10)/-2
  • y>3x + 5

Therefore, the answer is J.

Have a lovely rest of your day/night, and good luck with your assignments! ♡

3 0
2 years ago
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Find the x-intercept for 11x - 33y = 99
seropon [69]
It is 9. You have to add 33y, then divide by 11 and you get x = 3y + 9
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
Read 2 more answers
Am I correct? Please check my answer
avanturin [10]
Yes your right ...."...............................
6 0
3 years ago
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