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Dmitrij [34]
3 years ago
7

Find the line of inclination of the line if its slope is -√3​

Mathematics
1 answer:
gregori [183]3 years ago
4 0

Answer:

120° or 300°

Step-by-step explanation:

Given,

Slope (m) = -√3

Inclination (α) = ?

Now,

We know,

m = tanα

or, -√3 = tanα

or, tan(120° or 300°)  = tanα

∴ α = 120° or 300°

∴ Inclination is 120° or 300°

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Use the line. line it up line up the triangle
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Write the equation of the circle with center (0, 6) and radius 1.
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(x-h)^2 + (y-k)^2 = r^2

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The value (r) is the radius and is squared in the equation. In this case, the square root of 1 is 1. 

(x)^2 + (y-6)^2 = 1
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4 years ago
If x = y and y=z which statement is true?
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The x=y
Is this the the test of the question?
3 0
3 years ago
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What’s the correct answer for this?
Deffense [45]

Answer:

The first option

Step-by-step explanation:

In the image, we can notice that NQ is longer than PQ, similar to how JH is longer than NJ. Thus, NQ must be aligned with JH and PQ with JN. As PN and HN are clearly the longest for each triangle, those must be aligned as well.

All of the responses have NP and NH aligned.

The last response does not have NQ aligned with JH.

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6 0
4 years ago
Dwayne drove 18 miles to the airport to pick up his father and then returned home. On the return trip he was able to drive an av
Nesterboy [21]

Answer:

30 mph

Step-by-step explanation:

Let d = distance (in miles)

Let t = time (in hours)

Let v = average speed driving <u>to</u> the airport (in mph)

⇒ v + 15 = average speed driving <u>from</u> the airport (in mph)

Using:  distance = speed x time

\implies t=\dfrac{d}{v}

Create two equations for the journey to and from the airport, given that the distance one way is 18 miles:

\implies t=\dfrac{18}{v}  \ \ \textsf{and} \ \  t=\dfrac{18}{v+15}

We are told that the total driving time is 1 hour, so the sum of these expressions equals 1 hour:

\implies \dfrac{18}{v} +\dfrac{18}{v+15}=1

Now all we have to do is solve the equation for v:

\implies \dfrac{18(v+15)}{v(v+15)} +\dfrac{18v}{v(v+15)}=1

\implies \dfrac{18(v+15)+18v}{v(v+15)}=1

\implies 18(v+15)+18v=v(v+15)

\implies 18v+270+18v=v^2+15v

\implies v^2-21v-270=0

\implies (v-30)(v+9)=0

\implies v=30, v=-9

As v is positive, v = 30 only

So the average speed driving to the airport was 30 mph

(and the average speed driving from the airport was 45 mph)

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2 years ago
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