(04.05 MC)

Mathematics

1 answer:

nika2105 [10]3 years ago

6 0

Answer:

Step-by-step explanation:

Plug in and see.

Check A: Lets plug in (0,3) into y<2/3 x+2.

3<2/3 (0)+2

3<2 is not true so not A

Check C: Lets plug in (3,5) into y<2/3 x+2.

5<2/3 (3)+2

5<2+2

5<4 is not true so not C

Check B: Lets plug in (-3,1) into y<2/3 x+2.

1<2/3 (-3)+2

1<0 is not true so not B

Check D: Lets plug in (1,2) into y<2/3 x+2.

2<2/3 (1)+2

2<2/3+2 is true so D

You might be interested in

A = b + rb, for r Please help me..

kifflom [539]

Answer:

r=(a/b)-1

Step-by-step explanation:

a=b(1+r)

a/b = 1+r

r=(a/b)-1

6 0

3 years ago

A ball is thrown from an initial height of 3 meters with an initial upward velocity of 30 m/s. The ball's height h (in meters) a

yanalaym [24]

$h=3+30t-5t^2\implies 13=3+30t-5t^2\implies 0=-10+30t-5t^2 \\\\\\ 5t^2-30t+10=0\implies 5(t^2-6t+2)=0\implies t^2-6t+2=0$

now, it doesn't give us nice looking integers, so let's plug that into the quadratic formula.

$~~~~~~~~~~~~\textit{quadratic formula} \\\\ \stackrel{\stackrel{a}{\downarrow }}{1}t^2\stackrel{\stackrel{b}{\downarrow }}{-6}t\stackrel{\stackrel{c}{\downarrow }}{+2}=0 ~\hspace{10em} t= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a} \\\\\\ t= \cfrac{ - (-6) \pm \sqrt { (-6)^2 -4(1)(2)}}{2(1)}\implies t=\cfrac{6\pm\sqrt{36-8}}{2}$

$t=\cfrac{6\pm\sqrt{28}}{2}\implies t=\cfrac{6\pm\sqrt{2^2\cdot 7}}{2}\implies t=\cfrac{6\pm 2\sqrt{7}}{2}\implies t=\cfrac{2(3\pm 1\sqrt{7})}{2} \\\\\\ t=3\pm \sqrt{7}\implies t= \begin{cases} 3+\sqrt{7}\\ 3-\sqrt{7} \end{cases}\implies t\approx \begin{cases} 5.65\\ 0.35 \end{cases}$