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2 years ago

Solve: 5 + n > 2

Mathematics

1 answer:

tiny-mole [99]2 years ago

3 0

$\\ \sf\longmapsto 5+n>2$

We have to find n
Take 5 to right side
sign will be changed

$\\ \sf\longmapsto n>2-5$

Simplify

$\\ \sf\longmapsto n>-3$

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Arturiano [62]

Step-by-step explanation:

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2 years ago

A sphere and a cylinder have the same radius and height. The volume of the cylinder is 21m. what is the volume of the sphere.

Goshia [24]

Answer:

The volume of the sphere is 14m³

Step-by-step explanation:

Given

Volume of the cylinder = $21m^3$

Required

Volume of the sphere

Given that the volume of the cylinder is 21, the first step is to solve for the radius of the cylinder;

<em>Using the volume formula of a cylinder</em>

The formula goes thus

$V = \pi r^2h$

Substitute 21 for V; this gives

$21 = \pi r^2h$

Divide both sides by h

$\frac{21}{h} = \frac{\pi r^2h}{h}$

$\frac{21}{h} = \pi r^2$

The next step is to solve for the volume of the sphere using the following formula;

$V = \frac{4}{3}\pi r^3$

Divide both sides by r

$\frac{V}{r} = \frac{4}{3r}\pi r^3$

Expand Expression

$\frac{V}{r} = \frac{4}{3}\pi r^2$

Substitute $\frac{21}{h} = \pi r^2$

$\frac{V}{r} = \frac{4}{3} * \frac{21}{h}$

$\frac{V}{r} = \frac{84}{3h}$

$\frac{V}{r} = \frac{28}{h}$

Multiply both sided by r

$r * \frac{V}{r} = \frac{28}{h} * r$

$V = \frac{28r}{h}$ ------ equation 1

From the question, we were given that the height of the cylinder and the sphere have equal value;

This implies that the height of the cylinder equals the diameter of the sphere. In other words

$h = D$ , where D represents diameter of the sphere

Recall that $D = 2r$

So, $h = D = 2r$

$h = 2r$

Substitute 2r for h in equation 1

$V = \frac{28r}{2r}$

$V = \frac{28}{2}$

$V = 14$

Hence, the volume of the sphere is 14m³

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2 years ago

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2 years ago

2^-5/2^-6 PLEASE HELP <br> NO LINKS

emmasim [6.3K]

The answer is 1

explanation:
when dividing exponents you have to subtract them, -5 - (-6) is 1.
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2 years ago

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The rectangular bar is connected to the support bracket with a 10-mm-diameter pin. The bar width is w = 70 mm and the bar thickn

german

Answer:

P_max = 9.032 KN

Step-by-step explanation:

Given:

- Bar width and each side of bracket w = 70 mm

- Bar thickness and each side of bracket t = 20 mm

- Pin diameter d = 10 mm

- Average allowable bearing stress of (Bar and Bracket) T = 120 MPa

- Average allowable shear stress of pin S = 115 MPa

Find:

The maximum force P that the structure can support.

Solution:

- Bearing Stress in bar:

T = P / A

P = T*A

P = (120) * (0.07*0.02)

P = 168 KN

- Shear stress in pin:

S = P / A

P = S*A

P = (115)*pi*(0.01)^2 / 4

P = 9.032 KN

- Bearing Stress in each bracket:

T = P / 2*A

P = T*A*2

P = 2*(120) * (0.07*0.02)

P = 336 KN

- The maximum force P that this structure can support:

P_max = min (168 , 9.032 , 336)

P_max = 9.032 KN

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2 years ago