All categories

3 years ago

How do you find the number of degrees in a third angle of a triangle

Mathematics

1 answer:

Setler79 [48]3 years ago

3 0

every triangle must equal 180 degrees so add the other two angles and subtract by 180 to get that angle

Hope this helps :)

You might be interested in

Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by

11111nata11111 [884]

Answer:

The dimension of the open rectangular box is $8.216\times 4.216\times 1.392$ .

The volume of the box is 8.217 cubic inches.

Step-by-step explanation:

Given : The open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by cutting congruent squares from the corners and folding up the sides.

To find : The dimensions and the volume of the open rectangular box ?

Solution :

Let the height be 'x'.

The length of the box is '11-2x'.

The breadth of the box is '7-2x'.

The volume of the box is $V=l\times b\times h$

$V=(11-2x)\times (7-2x)\times x$

$V=4x^3-36x^2+77x$

Derivate w.r.t x,

$V'(x)=4(3x^2)-2(36x)+77$

$V'(x)=12x^2-72x+77$

The critical point when V'(x)=0

$12x^2-72x+77=0$

Solve by quadratic formula,

$x=\frac{18+\sqrt{93}}{6},\frac{18-\sqrt{93}}{6}$

$x=4.607,1.392$

Derivate again w.r.t x,

$V''(x)=24x-72$

Now, $V''(4.607)=24(4.607)-72=38.568>0$ (+ve)

$V''(1.392)=24(1.392)-72=-38.592$ (-ve)

So, there is maximum at x=1.392.

The length of the box is $l=11-2x$

$l=11-2(1.392)=8.216$

The breadth of the box is $b=7-2x$

$b=7-2(1.392)=4.216$

The height of the box is h=1.392.

The dimension of the open rectangular box is $8.216\times 4.216\times 1.392$ .

The volume of the box is $V=l\times b\times h$

$V=8.216\times 4.216\times 1.392$

$V=48.217\ in.^3$

The volume of the box is 8.217 cubic inches.

5 0

3 years ago

I need some really help......

kozerog [31]

Length x width = formula of finding the area of a rectangle

2 3/7 x 2 4/5 = 6 4/5 (6.8)

The area of the rectangle is 6 4/5 / 6.8

3 0

2 years ago

What is the answer? Please! ill mark brainliest!

Olin [163]

B, because the mid segment is half the length of the third side

4 0

2 years ago

Let ρ = x3 + xe−x for x ∈ (0, 1), compute the center of mass.

hram777 [196]

The center of mass is mathematically given as

$\bar{x}=\left(\frac{44 e-100}{25 e-40}\right)\end{aligned}$

<h3>What is the center of mass.?</h3>

Determine the center of mass in one dimension:

Represent the masses at the respective distances.

$\begin{|c|c|} Masses \ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ Located at \\\rho=x^{3}+x \cdot e^{-x} & \ \ \ \ x \in(0,1)$ \\\end$

We calculate the total mass of the system.

$\begin{aligned}m &=\int_{0}^{1} \rho \cdot d x \\& m =\int_{0}^{1}\left(x^{3}+x \cdot e^{-x}\right) \cdot d x \\&m =\left|\frac{x^{4}}{4}-(x+1) e^{-x}\right|_{0}^{1} \\&m =\left(\frac{5}{4}-\frac{2}{e}\right)\end{aligned}$

Step 03: Calculate the moment of the system.

$\begin{aligned}M &=\int_{0}^{1}(\rho \cdot x) \cdot d x \\& M=\int_{0}^{1}\left(x^{4}+x^{2} \cdot e^{-x}\right) \cdot d x \\&M =\left|\frac{x^{5}}{5}-\left(x^{2}-2 x+2\right) \cdot e^{-x}\right|_{0}^{1} \\&M=\left(\frac{11}{5}-\frac{5}{e}\right)\end{aligned}$

we calculate the center of mass.

$\begin{aligned}\bar{x} &=\left(\frac{M}{m}\right) \\& \bar{x}=\left\{\left(\frac{\left.11-\frac{5}{5}\right)}{\left(\frac{5}{4}-\frac{2}{e}\right)}\right\}\right.\\& \bar{x}=\left(\frac{11 e-25}{5 e}\right) \cdot\left(\frac{4 e}{5 e-8}\right) \\&\bar{x}=\left(\frac{44 e-100}{25 e-40}\right)\end{aligned}$