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Leokris [45]
3 years ago
15

You are asked to draw a triangle with side lengths of 6 inches and 8 inches. What is the longest whole number length that your t

hird side can be?
Mathematics
2 answers:
stepladder [879]3 years ago
4 0
The angle between the 6 inches and 8 inches length sides must be less than 180 degrees.
When the angle between the 6 inches and 8 inches length sides 180 degrees, the length of the third side will be 14 inches.
Therefore, the longest whole number length of the third side is 13 inches.
tensa zangetsu [6.8K]3 years ago
3 0
Fo sides, a,b,c
where c is the longest

c<a+b
or else you won't have a triangle

we want the longest side
this mystery side is c
other ones are a and b

c<6+8
c<14
the next largest whole number is 13

answer is 13 inches
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A solid is formed by adjoining two hemispheres to the ends of a right circular cylinder. An industrial tank of this shape must h
mestny [16]

Answer:

Radius =6.518 feet

Height = 26.074 feet

Step-by-step explanation:

The Volume of the Solid formed  = Volume of the two Hemisphere + Volume of the Cylinder

Volume of a Hemisphere  =\frac{2}{3}\pi r^3

Volume of a Cylinder =\pi r^2 h

Therefore:

The Volume of the Solid formed

=2(\frac{2}{3}\pi r^3)+\pi r^2 h\\\frac{4}{3}\pi r^3+\pi r^2 h=4640\\\pi r^2(\frac{4r}{3}+ h)=4640\\\frac{4r}{3}+ h =\frac{4640}{\pi r^2} \\h=\frac{4640}{\pi r^2}-\frac{4r}{3}

Area of the Hemisphere =2\pi r^2

Curved Surface Area of the Cylinder =2\pi rh

Total Surface Area=

2\pi r^2+2\pi r^2+2\pi rh\\=4\pi r^2+2\pi rh

Cost of the Hemispherical Ends  = 2 X  Cost of the surface area of the sides.

Therefore total Cost, C

=2(4\pi r^2)+2\pi rh\\C=8\pi r^2+2\pi rh

Recall: h=\frac{4640}{\pi r^2}-\frac{4r}{3}

Therefore:

C=8\pi r^2+2\pi r(\frac{4640}{\pi r^2}-\frac{4r}{3})\\C=8\pi r^2+\frac{9280}{r}-\frac{8\pi r^2}{3}\\C=\frac{9280}{r}+\frac{24\pi r^2-8\pi r^2}{3}\\C=\frac{9280}{r}+\frac{16\pi r^2}{3}\\C=\frac{27840+16\pi r^3}{3r}

The minimum cost occurs at the point where the derivative equals zero.

C^{'}=\frac{-27840+32\pi r^3}{3r^2}

When \:C^{'}=0

-27840+32\pi r^3=0\\27840=32\pi r^3\\r^3=27840 \div 32\pi=276.9296\\r=\sqrt[3]{276.9296} =6.518

Recall:

h=\frac{4640}{\pi r^2}-\frac{4r}{3}\\h=\frac{4640}{\pi*6.518^2}-\frac{4*6.518}{3}\\h=26.074 feet

Therefore, the dimensions that will minimize the cost are:

Radius =6.518 feet

Height = 26.074 feet

5 0
3 years ago
What can be used to determine the length of segment AB?
Vikki [24]

It depends on what you mean by "what can be used", you can use a protractor or ruler since it's a line segment. OR you can use the formula a^ + b^ = c^ to find out all the measurements.......

If you still can't find it, then, you can try to find the other measurements around that specific line segments. And it DEPENDS on what you choose to find the length of this segment.

( There's no picture on my screen, so I'm guessing that you didn't put any.. )

Well, I hope this can help you :3

STAY SAFE!! :)

6 0
3 years ago
bird that was perched atop a 15 ½ foot tree dives down six feet to a branch below. How far above the ground is the bird’s new lo
never [62]

Answer:

ok so if the bird is on the top at 15 and a half then goes down six feet substrate 15 and a half from six and you get nine and half from the ground to the bird

Step-by-step explanation:

8 0
3 years ago
1-SCPS
Anton [14]

Answer:

6

Step-by-step explanation:

5 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Cfrac%7Ba%5E%7B2%7D-1%7D%7B2-5a%7D%20times%20%5Cfrac%7B15a-6%7D%7Ba%5E%7B2%7D%2B5a-6%7D" id=
stepan [7]
\bf \cfrac{a^2-1}{2-5a}\times \cfrac{15-6}{a^2+5a-6}\\\\&#10;-----------------------------\\\\&#10;recall\quad \textit{difference of squares}&#10;\\ \quad \\&#10;(a-b)(a+b) = a^2-b^2\qquad \qquad &#10;a^2-b^2 = (a-b)(a+b)\\\\&#10;thus\quad a^2-1\iff a^2-1^2\implies (a-1)(a+1)&#10;\\\\\\&#10;now\quad a^2+5a-6\implies (a+6)(a-1)\\\\&#10;-----------------------------\\\\&#10;thus&#10;\\\\\\&#10;\cfrac{a^2-1}{2-5a}\times \cfrac{15-6}{a^2+5a-6}\implies \cfrac{(a-1)(a+1)}{2-5a}\times \cfrac{3(5a-2)}{(a+6)(a-1)}\\\\&#10;-----------------------------\\\\&#10;

\bf now\quad 3(5a-2) \iff -3(2-5a)\\\\&#10;-----------------------------\\\\&#10;thus&#10;\\\\\\&#10;\cfrac{\underline{(a-1)}(a+1)}{\underline{2-5a}}\times \cfrac{-3\underline{(2-5a)}}{(a+6)\underline{(a-1)}}\implies \cfrac{-3(a+1)}{a+6}
6 0
3 years ago
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