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2 years ago

Find the lateral area of a cylinder that has a diameter of 18 mm and a height of 15 mm.

Mathematics

1 answer:

nika2105 [10]2 years ago

4 0

Answer:

848.23 mm²

Step-by-step explanation:

LA = πdh

LA = 3.14159 × 18 mm × 15 mm

LA = 848.23 mm²

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Simplify each rational expression to lowest terms, specifying the values of xx that must be excluded to avoid division

k0ka [10]

Answer:

(a) $\frac{x^2-6x+5}{x^2-3x-10}=\frac{x-1}{x+2}$ . The domain of this function is all real numbers not equal to -2 or 5.

(b) $\frac{x^3+3x^2+3x+1}{x^3+2x^2-x}=1+\frac{x^2+4x+1}{x^3+2x^2-x}$ . The domain of this function is all real numbers not equal to 0, $-1+\sqrt{2}$ or $-1+\sqrt{2}$ .

(c) $\frac{x^2-16}{x^2+2x-8}=\frac{x-4}{x-2}$ .The domain of this function is all real numbers not equal to 2 or -4.

(d) $\frac{x^2-3x-10}{x^3+6x^2+12x+8}=\frac{x-5}{\left(x+2\right)^2}$ . The domain of this function is all real numbers not equal to -2.

(e) $\frac{x^3+1}{x^2+1}=x+\frac{-x+1}{x^2+1}$ . The domain of this function is all real numbers.

Step-by-step explanation:

To reduce each rational expression to lowest terms you must:

(a) For $\frac{x^2-6x+5}{x^2-3x-10}$

$\mathrm{Factor}\:x^2-6x+5\\\\x^2-6x+5=\left(x^2-x\right)+\left(-5x+5\right)\\x^2-6x+5=x\left(x-1\right)-5\left(x-1\right)\\\\\mathrm{Factor\:out\:common\:term\:}x-1\\x^2-6x+5=\left(x-1\right)\left(x-5\right)$

$\mathrm{Factor}\:x^2-3x-10\\\\x^2-3x-10=\left(x^2+2x\right)+\left(-5x-10\right)\\x^2-3x-10=x\left(x+2\right)-5\left(x+2\right)\\\\\mathrm{Factor\:out\:common\:term\:}x+2\\x^2-3x-10=\left(x+2\right)\left(x-5\right)$

$\frac{x^2-6x+5}{x^2-3x-10}=\frac{\left(x-1\right)\left(x-5\right)}{\left(x+2\right)\left(x-5\right)}$

$\mathrm{Cancel\:the\:common\:factor:}\:x-5\\\\\frac{x^2-6x+5}{x^2-3x-10}=\frac{x-1}{x+2}$

The denominator in a fraction cannot be zero because division by zero is undefined. So we need to figure out what values of the variable(s) in the expression would make the denominator equal zero.

To find any values for x that would make the denominator = 0 you need to set the denominator = 0 and solving the equation.

$x^2-3x-10=\left(x+2\right)\left(x-5\right)=0$

Using the Zero Factor Theorem: = 0 if and only if = 0 or = 0

$x+2=0\\x=-2\\\\x-5=0\\x=5$

The domain is the set of all possible inputs of a function which allow the function to work. Therefore the domain of this function is all real numbers not equal to -2 or 5.

(b) For $\frac{x^3+3x^2+3x+1}{x^3+2x^2-x}$

$\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}x^3+3x^2+3x+1\mathrm{\:and\:the\:divisor\:}x^3+2x^2-x\mathrm{\::\:}\frac{x^3}{x^3}=1$

Quotient = 1

$\mathrm{Multiply\:}x^3+2x^2-x\mathrm{\:by\:}1:\:x^3+2x^2-x$

$\mathrm{Subtract\:}x^3+2x^2-x\mathrm{\:from\:}x^3+3x^2+3x+1\mathrm{\:to\:get\:new\:remainder}$

Remainder = $x^2+4x+1}$

$\frac{x^3+3x^2+3x+1}{x^3+2x^2-x}=1+\frac{x^2+4x+1}{x^3+2x^2-x}$

The domain of this function is all real numbers not equal to 0, $-1+\sqrt{2}$ or $-1+\sqrt{2}$ .

$x^3+2x^2-x=0\\\\x^3+2x^2-x=x\left(x^2+2x-1\right)=0\\\\\mathrm{Solve\:}\:x^2+2x-1=0:\quad x=-1+\sqrt{2},\:x=-1-\sqrt{2}$

(c) For $\frac{x^2-16}{x^2+2x-8}$

$x^2-16=\left(x+4\right)\left(x-4\right)$

$x^2+2x-8= \left(x-2\right)\left(x+4\right)$

$\frac{x^2-16}{x^2+2x-8}=\frac{\left(x+4\right)\left(x-4\right)}{\left(x-2\right)\left(x+4\right)}\\\\\frac{x^2-16}{x^2+2x-8}=\frac{x-4}{x-2}$

The domain of this function is all real numbers not equal to 2 or -4.

$x^2+2x-8=0\\\\x^2+2x-8=\left(x-2\right)\left(x+4\right)=0$

(d) For $\frac{x^2-3x-10}{x^3+6x^2+12x+8}$

$\mathrm{Factor}\:x^2-3x-10\\\left(x^2+2x\right)+\left(-5x-10\right)\\x\left(x+2\right)-5\left(x+2\right)$

$\mathrm{Apply\:cube\:of\:sum\:rule:\:}a^3+3a^2b+3ab^2+b^3=\left(a+b\right)^3\\\\a=x,\:\:b=2\\\\x^3+6x^2+12x+8=\left(x+2\right)^3$

$\frac{x^2-3x-10}{x^3+6x^2+12x+8}=\frac{\left(x+2\right)\left(x-5\right)}{\left(x+2\right)^3}\\\\\frac{x^2-3x-10}{x^3+6x^2+12x+8}=\frac{x-5}{\left(x+2\right)^2}$

The domain of this function is all real numbers not equal to -2

$x^3+6x^2+12x+8=0\\\\x^3+6x^2+12x+8=\left(x+2\right)^3=0\\x=-2$

(e) For $\frac{x^3+1}{x^2+1}$

$\frac{x^3+1}{x^2+1}=x+\frac{-x+1}{x^2+1}$

The domain of this function is all real numbers.

$x^2+1=0\\x^2=-1\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{-1},\:x=-\sqrt{-1}$

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3 years ago

Can some one help me with 1,3,and 5

FrozenT [24]

1. 2/100 or 1/50, as 0.02 is equal to 2 hundreths. Simplifying gets you 1/50

3. 1/2 or 50/100, as 0.50 is equal to 0.5 which is equal to 50 hundreths. Simplify to get 1/2

5. 88/100 or 44/50 , as 0.88 is equal to 88 hundreths. After simplifying, you get 44/50

Depending on your teacher, you may be told to not simplify. It depends

8 0

3 years ago

Find the lateral area of a cylinder that has a diameter of 18 mm and a height of 15 mm.​

Find the lateral area of a cylinder that has a diameter of 18 mm and a height of 15 mm.