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3 years ago

YOOO MARKIN BRAINIEST !!

Mathematics

2 answers:

Mkey [24]3 years ago

5 0

Answer:

Step-by-step explanation:

Brilliant_brown [7]3 years ago

3 0

Y ≥ 2x-5
y-intercept is -5, from there go 2 up 1 over ( slope is rise over run)
greater than or equal so the line is connected
now find the shaded area by plugging in (0,0)
0 ≥2(0)-5
0 ≥-5 is correct so shade the area that to the left of the line, (the whole area that including (0,0))

y<-3x
y intercept is 0 so start from there and go down 3 right 1 (or go up 3 left 1)
broken like cause no or equal sign
the (0,0) is on the line so use (1,1) to find the answer
1<-3(1) is incorrect so shade the area that dies not include (1,1) or the entire area to the left of that line

you can see the section where both shaded area cross, thats the answer so erase every area you shaded that isn’t the answer so
THE ANSWER IS C

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2 years ago

Please help me i need an answer

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2 years ago

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2 years ago

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The following statement is either true (in all cases) or false (for at least one example). If false, construct a specific e

matrenka [14]

Answer:

False, counterexample below

Step-by-step explanation:

Denote the unit vectors of R^3 by $e_1=(1,0,0), e_2=(0,1,0), e_3=(0,0,1)$ . Now consider $v_1=e_1, v_2=2e_1$ and $v_3=e_3$ . We have that $v_1, v_2,v_3 \in \mathbb{R}^3$ . Also, the vector $v_3$ is not a linear combination of $v_1, v_2$ because any linear combination of these two vectors will have third coordinate zero, but v_3 has third coordinate 1 so they can't be equal.

However, the set $\{v_1, v_2,v_3\}$ is not linearly independent, because $2v_1-v_2+0v_3=0$ is a non-trivial linear combination of these vectors that equals zero.

4 0

3 years ago

A parabola can be drawn given a focus of (-11, -2) and a directrix of x= -3

fgiga [73]

Check the picture below, so the parabola looks more or less like that.

now, the vertex is half-way between the focus point and the directrix, so that puts it where you see it in the picture, and the horizontal parabola is opening to the left-hand-side, meaning that the distance "P" is negative.

$\textit{horizontal parabola vertex form with focus point distance} \\\\ 4p(x- h)=(y- k)^2 \qquad \begin{cases} \stackrel{vertex}{(h,k)}\qquad \stackrel{focus~point}{(h+p,k)}\qquad \stackrel{directrix}{x=h-p}\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix}\\\\ \stackrel{"p"~is~negative}{op ens~\supset}\qquad \stackrel{"p"~is~positive}{op ens~\subset} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}$

$\begin{cases} h=-7\\ k=-2\\ p=-4 \end{cases}\implies 4(-4)[x-(-7)]~~ = ~~[y-(-2)]^2 \\\\\\ -16(x+7)=(y+2)^2\implies x+7=-\cfrac{(y+2)^2}{16}\implies x=-\cfrac{1}{16}(y+2)^2-7$

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2 years ago