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lora16 [44]
2 years ago
13

What is the slope between y=3.20x+9 simplified

Mathematics
1 answer:
sattari [20]2 years ago
6 0

Answer:

3.20

Step-by-step explanation:

Intercept form of a line : Y = mx + b

m = y2 - y2 / x2 - x1 ....use this to find the slope whenever 2 points are given.

In this case it's directly given.

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Kathy swam 3 laps in the pool this week. She must swim more than 12 laps.
Marysya12 [62]

Answer:10

Step-by-step explanation:

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3 years ago
Suppose that the number of drivers who travel between a particular origin and destination during a designated time period has a
andrew11 [14]

Answer:

Step-by-step explanation:

Given that the number of drivers who travel between a particular origin and destination during a designated time period has a Poisson distribution with parameter μ = 20 (suggested in the article "Dynamic Ride Sharing: Theory and Practice"†).

a) P(X\leq 15) = 0.1565=0.157

b) P(X>26) =1-F(26)\\= 1-0.9221\\=0.0779=0.078

c) P(15\leq x\leq 26)\\=F(26)-F(14)\\=0.9221-0.1049\\=0.8172=0.817

d) 2 std dev = 2(20) =40

Hence 2 std deviation means

20-40, 20+40

i.e. (0,60)

P(0

5 0
3 years ago
(15 ^3) x 15^-6 <br> Plz help me
masya89 [10]

put 3 into 15 than put 6 into 15 then the answer u get from putting 6 into 15 subtract that by 2 and the answer u get from the u multpy the answer that u got from putting 3 into 15 together

4 0
3 years ago
If the Circumference of a circle is 12 cm, what is the circles Diameter? Please help thankss!
BARSIC [14]

Answer:

Step-by-step explanation:

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5 0
3 years ago
Simplify the expression. tan(sin^−1 x)
Blizzard [7]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2799412

_______________


Let  \mathsf{\theta=sin^{-1}(x)\qquad\qquad-\dfrac{\pi}{2}\ \textless \ \theta\ \textless \ \dfrac{\pi}{2}.}

(that is the range of the inverse sine function).


So,

\mathsf{sin\,\theta=sin\!\left[sin^{-1}(x)\right]}\\\\ \mathsf{sin\,\theta=x\qquad\quad(i)}


Square both sides:

\mathsf{sin^2\,\theta=x^2\qquad\qquad(but~sin^2\,\theta=1-cos^2\,\theta)}\\\\ \mathsf{1-cos^2\,\theta=x^2}\\\\ \mathsf{1-x^2=cos^2\,\theta}\\\\ \mathsf{cos^2\,\theta=1-x^2}


Since \mathsf{-\,\dfrac{\pi}{2}\ \textless \ \theta\ \textless \ \dfrac{\pi}{2},} then \mathsf{cos\,\theta} is positive. So take the positive square root and you get

\mathsf{cos\,\theta=\sqrt{1-x^2}\qquad\quad(ii)}


Then,

\mathsf{tan\,\theta=\dfrac{sin\,\theta}{cos\,\theta}}\\\\\\ \mathsf{tan\,\theta=\dfrac{x}{\sqrt{1-x^2}}}\\\\\\\\ \therefore~~\mathsf{tan\!\left[sin^{-1}(x)\right]=\dfrac{x}{\sqrt{1-x^2}}\qquad\qquad -1\ \textless \ x\ \textless \ 1.}


I hope this helps. =)


Tags:  <em>inverse trigonometric function sin tan arcsin trigonometry</em>

3 0
3 years ago
Read 2 more answers
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