Find the area of the figure.

Mathematics

2 answers:

posledela3 years ago

7 0

Answer:

29 sq in

Step-by-step explanation:

Area of the figure = Area of square + Area of parallelogram + Area of rectangle

$= {2}^{2} + 5 \times 3 + 2 \times 5 \\ \\ = 4 + 15 + 10 \\ \\ = 29 \: {in}^{2}$

Klio2033 [76]3 years ago

4 0

Answer:

29 in

Step-by-step explanation:

the rectangle at the bottom would be 10 when multiplying its base and its height which are 5 and 2.

the parallelogram in the middle would be 15 from multiplying 5 * 3.

and the square at the top would be 4 from multiplying 2 * 2.

so now you add that up and you should get the total i got.

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G(x) = -2x^3 – 15x^2 + 36x

shusha [124]

Consider the function $G(x) = -2x^3 - 15x^2 + 36x.$ First, factor it:

$G(x) = -2x^3 - 15x^2 + 36x=-x(2x^2+15x-36)=\\ \\=-x\cdot 2\cdot \left(x-\dfrac{-15-\sqrt{513}}{4}\right)\cdot \left(x-\dfrac{-15+\sqrt{513}}{4}\right).$

The x-intercepts are at points $\left(\dfrac{-15-\sqrt{513} }{4},0\right),\ (0,0),\ \left(\dfrac{-15+\sqrt{513} }{4},0\right).$

1. From the attached graph you can see that

function is positive for $x\in \left(-\infrty, \dfrac{-15-\sqrt{513} }{4}\right)\cup \left(0,\dfrac{-15+\sqrt{513} }{4}\right);$
function is negative for $x\in \left(\dfrac{-15-\sqrt{513} }{4},0\right)\cup \left(\dfrac{-15+\sqrt{513} }{4},\infty\right).$