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3 years ago

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

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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! ♡

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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

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

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3 years ago

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Am I correct? Please check my answer

avanturin [10]

Yes your right ...."...............................

6 0

3 years ago