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expeople1 [14]
3 years ago
9

Can someone please explain Linear Inequalities to me?​

Mathematics
2 answers:
Arisa [49]3 years ago
8 0

A linear inequality is similar to a linear equation ( y=2x+5) where there is a relation between 2 variables and when plotted on the graph it is always a straight line, taking the example given earlier, y=2x+5 when plotted on a graph would give us this line ( added as attachment), a linear equations or inequalities do not have a power above, like x^2+3x+9=y is not a linear equation as there is a power above 1, in this case 2 involved, coming to linear inequalities, only the '=' symbol is replaced with \geq, \leq,. so instead of y=2x+5, a linear <u>inequality</u> would be :-

y\leq 2x+5\\ y\geq 2x+5\\ y2x+5

Now all of these have seperate meanings, starting with the first one it states y is less than or equal to 2x+5, second is y is greater than or equal to 2x+5, third is y is less than 2x+5 and lastly the fourth is y is greater than 2x+5. All 4 of these inequalities are graphed on the graph similarly to the original equation ( y=2x+5) however shading the right area according to the inquality is important.

In the graphs attached below, there is a solid line for \leq,\geq but a dashed line for <,> this is because the value of y in the first 2 equations may be equal to 2x+5, it may be on the line or any where in the shaded regions, however in the second too it cannot be on the line but it can only be in the shaded region hence the line is dashed

I hope I made sense for you, let me know if anything confused it, i'll elaborate it further

Zarrin [17]3 years ago
6 0
Ok so an equation like y<_x+2 is not completely equal. a linear inequality is like that but just graphed, so basically a graphed unequal equation (<_ means equal to or greater) i hope this helped!
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Dennis_Churaev [7]

Answer:

It would take approximately 6.50 second for the cannonball to strike the ground.

Step-by-step explanation:

Consider the provided function.

h(t)=-4.9t^2+30.5t+8.8

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Substitute h(t) = 0 in above function.

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Multiply both sides by 10.

-49t^2+305t+88=0

For a quadratic equation of the form ax^2+bx+c=0 the solutions are: x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

Substitute a = -49, b = 305 and c=88

t=\frac{-305+\sqrt{305^2-4\left(-49\right)88}}{2\left(-49\right)}=-\frac{-305+\sqrt{110273}}{98}\\t = \frac{-305-\sqrt{305^2-4\left(-49\right)88}}{2\left(-49\right)}= \frac{305+\sqrt{110273}}{98}

Ignore the negative value of t as time can't be a negative number.

Thus,

t=\frac{305+\sqrt{110273}}{98}\approx6.50

Hence, it would take approximately 6.50 second for the cannonball to strike the ground.

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