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1 year ago

Which graph shows the solution to the system of linear inequalities?

Mathematics

1 answer:

snow_tiger [21]1 year ago

6 0

The solution to the inequality y > x + 3 and y < -x + 2 is the dark green region shown in the graph.

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

Inequality shows the non equal comparison of two or more numbers and variables.

Given the inequality y > x + 3 and y < -x + 2.

The solution to the inequality y > x + 3 and y < -x + 2 is the dark green region shown in the graph.

Find out more on equation at: brainly.com/question/2972832

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The linearization or linear approximation of a function $f(x)$ is given by:

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$V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh$

Taking $R_0=1.5$ inches and $h=3$ inches and evaluating the partial derivatives we obtain:

$V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh\\V(R, h) = V(R_0, h_0) + (\frac{2 h \pi r}{3} + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh$

substituting the values and taking $dx=0.1$ and $dh=0.3$ inches we have:

$V(R_0+dR, h_0+dh) =V(R_0, h_0) + (\frac{2 h \pi r}{3} + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh\\V(1.5+0.1, 3+0.3) =V(1.5, 3) + (\frac{2 \cdot 3 \pi \cdot 1.5}{3} + 2 \pi 1.5^2)\cdot 0.1 + (\frac{\pi 1.5^2}{3} )\cdot 0.3\\V(1.5+0.1, 3+0.3) = 17.2002\\\boxed{V(1.5+0.1, 3+0.3) \approx 17.20}$

Therefore the change in volume is estimated to be 17.20 $\rm{in^3}$

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