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Ede4ka [16]
3 years ago
8

Which expression is NOT equivalent to the other three?

Mathematics
2 answers:
olasank [31]3 years ago
6 0
The answer is -6 + n + 9n
abruzzese [7]3 years ago
4 0
The answer is -6 + n + 9n
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I need help on this asap please. <br> im being timed.<br> and i put in a photo of the problem.
Elden [556K]

F=ma\\\\a.\\ma=F\qquad|\text{divide both sides by}\ a\neq0\\\\m=\dfrac{F}{a}\\\\b.\\F=-30,\ a=6.\ \text{Substitute}\\\\m=\dfrac{-30}{6}=-5\\\\c.\\ma=F\qquad|\text{divide both sides by}\ m\neq0\\\\a=\dfrac{F}{m}\\\\F=30,\ m=10.\text{Substitute}\\\\a=\dfrac{30}{10}=3

6 0
3 years ago
a coin is flipped and a die is rolled what is the probability of rolling a number greater than or equal to 2 on the die?​
laiz [17]
5/6
Coin flip is irrelevant to result of die
5 0
4 years ago
What is the slope of the line with the coordinates (0,0) and (-4,7).
Citrus2011 [14]

Answer:

Slope = -7/4

Step-by-step explanation:

Let's find the slope between your two points.

(0,0) ; (−4,7)

(x1,y1) = (0,0)

(x2,y2) = (−4,7)

Use the slope formula:

m = y2−y1 / x2−x1

= 7−0/−4−0

=7/−4

= −7/4

Answer:

m= −7/4

Hope this helps!!

3 0
4 years ago
PLEASE HELP IM STUCK
Eduardwww [97]

Answer:

sorry I do not know it you can try other apps

3 0
3 years ago
The focus of a parabola is (-3,-5) The directrix of the parabola is y=2
cricket20 [7]

Check the picture below.

so the focus point is there, and the directrix is above it, meaning is a vertical parabola and is opening downwards, since the parabola opens up towards the focus.

now, the vertex is half-way between those two guys, at a "p" distance from either one, if we move over the y-axis from -5 to +2, we have 7 units, half-way is 3.5 units, and that puts us at -1.5 or -1½, as you see in the picture, so the vertex is then at (-3 , -1½).

so the distance from the vertex to the focus point  is then 3½ units, however since the parabola is opening downwards, "p" is negative, thus "p = 3½".

\bf \textit{parabola vertex form with focus point distance} \\\\ \begin{array}{llll} 4p(x- h)=(y- k)^2 \\\\ \stackrel{\textit{using this one}}{4p(y- k)=(x- h)^2} \end{array} \qquad \begin{array}{llll} vertex\ ( h, k)\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}

\bf \begin{cases} h=-3\\ k=-\frac{3}{2}\\[0.7em] p=-\frac{7}{2} \end{cases}\implies 4\left( -\cfrac{7}{2} \right)\left[ y-\left(-\cfrac{3}{2} \right) \right]=\left[ x-\left( -3 \right) \right]^2 \\\\\\ -14\left( y+\cfrac{3}{2} \right)=(x+3)^2\implies y+\cfrac{3}{2} =-\cfrac{(x+3)^2}{14} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill y=-\cfrac{1}{14}(x+3)^2-\cfrac{3}{2}~\hfill

4 0
3 years ago
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