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Elden [556K]
3 years ago
11

INEQUALITY If -5 > a and a > b then -5 is * ​

Mathematics
1 answer:
LUCKY_DIMON [66]3 years ago
6 0

B is < -5

I hope this helps

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Solve 3x² + 4x = 2. (2 points)
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Answer:

Step-by-step explanation:

As it is a second order equation, it means that it has two possible answers and they are x_1 and x_2.

The famous quadratic formula for solving any second order equation is the following:

x_{1} = \frac{-b + \sqrt{b^2 - 4 ac}}{2a}\\x_{2}=\frac{-b - \sqrt{b^2 - 4 ac}}{2a}

Where a is the coefficient of x^2, b is the coefficient of x, and c is the free term. In other words,

a = 3\\b=4\\c=-2

as the equation should be in the following form:

a x^2 + bx+c = 0

Therefore the possible answer should be the following,

\frac{-4 + \sqrt{4^2 - 4*3*(-2)}}{2*3}=\frac{-4 +\sqrt{16 + 24}}{6} =\frac{-4 + \sqrt{40}}{6}=\\ \frac{-4 + \sqrt{4*10}}{6} = \frac{-4 + 2\sqrt{10}}{6}\frac{2*(-2 + \sqrt{10})}{2*3} = \frac{-2 + \sqrt{10}}{3}

by dividing the numerator and denominator by 2, we can deduce the following,

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2 years ago
Bob can dig holes and plant 15 trees in 60 minutes. George can accomplish this same task in 120 minutes. If they work together,
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A square room has a tiled floor with 81 square tiles. How many tiles are along an edge of the room?
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In a free-fall experiment, an object is dropped from a height of h = 400 feet. A camera on the ground 500 ft from the point of i
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the only force acting on the object, is gravity, using feet will then be -32ft/s²,


was wondering myself on -32 or 32.. but anyhow... we'll settle for the negative value, since it seems to be just a bit of convention issues

so, we'll do the integral to get v(t) then

\bf \displaystyle \int -32\cdot dt\implies -32t+C&#10;\\\\\\&#10;\textit{object moves from \underline{rest}, so velocity is 0 at 0secs}&#10;\\\\\\&#10;-32(0)+C=0\implies C=0\implies \boxed{v(t)=-32t}&#10;\\\\\\&#10;\textit{now to get the positional s(t)}&#10;\\\\\\&#10;\displaystyle \int -32t\cdot dt\implies -16t^2+C&#10;\\\\\\&#10;\textit{the initial \underline{position} was 400ft away at 0secs}&#10;\\\\\\&#10;-16(0)^2+C=400\implies C=400\implies \boxed{s(t)=-16t^2+400}

when will it reach the ground level? let's set s(t) = 0

\bf s(t)=-16t^2+400\implies 0=-16t^2+400\implies \cfrac{-400}{-16}=t^2&#10;\\\\\\&#10;25=t^2\implies \boxed{5=t}


part B)  check the picture below

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