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stira [4]
4 years ago
14

I need help with this question. It's about inequalities. ​

Mathematics
1 answer:
PtichkaEL [24]4 years ago
5 0
N is greater than it equal to 3

So you put a shaded in for a three and draw a line to the right


I think :)
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PLEASE URGENT HELP!!!
olchik [2.2K]

Answer:

The equation is;

y = 1/3x

Step-by-step explanation:

The equation of a linear model is given as;

y = mx + b

m is the slope while b is the y intercept

As we can see, the y intercept is the origin and that means b is zero

To get the value of m, the slope, we have to use the slope equation

We simply select any two points on the line

We have the points as (6,2) and (15,5)

The equation of the slope is:

m = (y2-y1)/(x2-x1)

(x1,y1) = (6,2)

(x2,y2) = (15,5)

m = (5-2)/(15-6) = 3/9 = 1/3

So the equation will be;

y = 1/3x

3 0
3 years ago
Determine whether or not the following graph represents two quantities that are proportional to each other. Explain your reasoni
m_a_m_a [10]

Answer:

Not proportional.

Step-by-step explanation:

The values do not begin from a straight line at the origin. X begins at 2, not the origin.

4 0
3 years ago
an online video game tournament begins with 729. two players in each game there is only one winner and only the winner advances
Dimas [21]

Answer:

729 games

i think

8 0
3 years ago
Read 2 more answers
Look at the attachment<br>Simplified it too <br>pls take it seriously ​
Phantasy [73]

\\ \rm\rightarrowtail \dfrac{sin(180+A)+2cos(180+A).cos(A-180)}{2cos^2(360+A)-cos(-A)}

\\ \rm\rightarrowtail \dfrac{-sinA+2(cos^2A-sin^2A)}{2cos^2A-cosA}

\\ \rm\rightarrowtail \dfrac{-sinA+2cos^2A-2sin^2A}{cosA(2cosA-1)}

\\ \rm\rightarrowtail \dfrac{-sinA+2-2sin^2A-2sin^2A}{cosA(2cosA-1)}

\\ \rm\rightarrowtail \dfrac{-sinA-4sin^2A}{cosA(2cosA-1)}

\\ \rm\rightarrowtail \dfrac{-sinA(4sinA-1)}{cosA(2cosA-1)}

\\ \rm\rightarrowtail -tanA\left(\dfrac{4sinA-1}{2cosA-1}\right)

5 0
2 years ago
Please solve: x^2+4x-8=0
Bezzdna [24]
x^2+4x-8=0\\\\\underbrace{x^2+2x\times2+2^2}_{use:(a+b)^2=a^2+2ab+b^2}-2^2-8=0\\\\(x+2)^2-4-8=0\\\\(x+2)^2-12=0\\\\(x+2)^2=12\iff x+2=\sqrt{12}\ or\ x+2=-\sqrt{12}\\\\x=\sqrt{4\times3}-2\ or\ x=-\sqrt{4\times3}-2\\\\x=\sqrt4\times\sqrt3-2\ or\ x=-\sqrt4\times\sqrt3-2\\\\\center\boxed{x=2\sqrt3-2\ or\ x=-2\sqrt3-2}
4 0
3 years ago
Read 2 more answers
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