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Rainbow [258]
3 years ago
10

Write an equation that describes the function.

Mathematics
2 answers:
vlabodo [156]3 years ago
7 0
Find the slope, first:
m =  \frac{ y_{1}  - y_{2} }{ x_{1}  - x_{2} }
Use any two points, for example (1,0) and (3, -2)
m =  \frac{-2 - 0}{3 - 1}  = -1
So, the slope is -1.
Then, use the formula for any given point, y and x are constants
y -  y_{1}  = m(x -  x_{1} )
Use, for example, point (1, 0)
y - 0 = (-1)(x - 1)
y = -x +1
So, the equation is y = -x +1
Leona [35]3 years ago
5 0
From the points you get a slope of -1 using the slope in y=mx+b you get b=1 so y=-x+1
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Lelechka [254]

Answer:

1) 14/3 (D)

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Step-by-step explanation:

1) 5-(2/3)*(2/3//4/3)

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3 years ago
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3 years ago
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Bumek [7]
First thing First. You must find have a common denominator. If you want to find it then you do this

3: 3 6 9 12 15 18 21
7: 7 14 21 28

Once you found a number they have in common (21) You do this next

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3 0
3 years ago
Please help me with this assignment ASAP!!! I only partially understand, so please explain. Also it's a work sample.
Rudiy27
To find the answer, we can first find the distance each of them have, then divide it by their speed to find the time needed.

Distance Steve need to travel: 300 feet
Distance Paula need to travel:
300+175(behind steve) = 475 feet

Time needed respectively:

Steve: 300 ÷ 9 = 33.33.. seconds
Paula: 475 ÷ 15 = 31.66... seconds

As we can see from the result, Paula would take less time to reach the goal (31.66<33.33),
therefore, Paula win the bike race by:

33.33-31.67
=1.66...
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Thus Paula won by 1.67 feet.

Hope it helps!
5 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
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