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svp [43]
3 years ago
7

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 :)


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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}

Read more about the center of mass.

brainly.com/question/27549055

#SPJ1

8 0
2 years ago
On a coordinate plane the location of 3 points are (0,6) (5,-4) and (-1,-5) where would the fourth point need to be located in o
REY [17]

Answer:

(9.2)

Step-by-step explanation:

idkc

5 0
2 years ago
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