All categories

3 years ago

Find the area of the composite figure

Mathematics

1 answer:

Alja [10]3 years ago

7 0

Answer:

37 sq. ft.

Step-by-step explanation:

Multiply 1.5 by 2 and add it to 4 to get the bottom of the triangle part, which is 7. The. Multiply it by 6 and divide by 2 to get the area of the triangle, which is 21. Then find the area of the square part, which is 16, and finally add both areas together to get the total area of the composite figure

You might be interested in

1. [-/1 Points]

stiv31 [10]

multiply x times y divide by 6 on both sides and you got your answer

4 0

9 months ago

For the data in the table, write an equation for the direct variation.

melisa1 [442]

The equation of direct variation is,

$\begin{gathered} y=kx \\ k=\frac{y}{x} \end{gathered}$

Substitute the values of y and x.

$\begin{gathered} k=\frac{5.4}{4} \\ k=1.35 \end{gathered}$

So, the obtained equation will be y=1.35(x).

Let's check with the values of x.

1) If x=4

$\begin{gathered} y=1.35x \\ y=1.35(4) \\ y=5.4 \end{gathered}$

2) If x=2

$\begin{gathered} y=1.35x \\ y=1.35(2) \\ y=2.7 \end{gathered}$

3) If x=-2

4 0

6 months ago

Let S be the solid beneath z = 12xy^2 and above z = 0, over the rectangle [0, 1] × [0, 1]. Find the value of m > 1 so that th

jonny [76]

Answer:

The answer is $\sqrt{\frac{6}{5}}$

Step-by-step explanation:

To calculate the volumen of the solid we solve the next double integral:

$\int\limits^1_0\int\limits^1_0 {12xy^{2} } \, dxdy$

Solving:

$\int\limits^1_0 {12x} \, dx \int\limits^1_0 {y^{2} } \, dy$

$[6x^{2} ]{{1} \atop {0}} \right. * [\frac{y^{3}}{3}]{{1} \atop {0}} \right.$

Replacing the limits:

$6*\frac{1}{3} =2$

The plane y=mx divides this volume in two equal parts. So volume of one part is 1.

Since m > 1, hence mx ≤ y ≤ 1, 0 ≤ x ≤ $\frac{1}{m}$

Solving the double integral with these new limits we have:

$\int\limits^\frac{1}{m} _0\int\limits^{1}_{mx} {12xy^{2} } \, dxdy$

This part is a little bit tricky so let's solve the integral first for dy:

$\int\limits^\frac{1}{m}_0 [{12x \frac{y^{3}}{3}}]{{1} \atop {mx}} \right.\, dx =\int\limits^\frac{1}{m}_0 [{4x y^{3 }]{{1} \atop {mx}} \right.\, dx$

Replacing the limits:

$\int\limits^\frac{1}{m}_0 {4x(1-(mx)^{3} )\, dx =\int\limits^\frac{1}{m}_0 {4x-4x(m^{3} x^{3} )\, dx =\int\limits^\frac{1}{m}_0 ({4x-4m^{3} x^{4}) \, dx$

Solving now for dx:

$[{\frac{4x^{2}}{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right. = [{2x^{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right.$

Replacing the limits:

$\frac{2}{m^{2} }-\frac{4m^{3}\frac{1}{m^{5}}}{5} =\frac{2}{m^{2} }-\frac{4\frac{1}{m^{2}}}{5} \\ \frac{2}{m^{2} }-\frac{4}{5m^{2} }=\frac{10m^{2}-4m^{2} }{5m^{4}} \\ \frac{6m^{2} }{5m^{4}} =\frac{6}{5m^{2}}$