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lesantik [10]
4 years ago
13

Please answer this question! Thank you!

Mathematics
1 answer:
Dahasolnce [82]4 years ago
8 0
Simple...

you have: -x+2y=8

To find the y-intercept...re-write in y=mx+b form-->>>

-x+2y=8

-x+2y=8
+x      +x

2y=x+8

Divide by 2....

\frac{2y}{2} = \frac{x+8}{2}

y=\frac{x}{2} +4

Your y-intercept is 4. (0,4)

Thus, your answer.
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The substraction of a matrix B may be considered as the addition of the matrix (-1)B. Dose the commutative law of addition permi
GuDViN [60]

Answer:  The corrected statement is A - B = -B + A.

Step-by-step explanation:  Given that the subtraction of a matrix B may be considered as the addition of the matrix (-1)B.

We are given to check whether the commutative law of addition permit us to state that A - B = B - A.

If not, We are to correct the statement.

If the subtraction A - B is considered a the addition A + (-B), then the commutative law should be stated as follows :

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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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