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3 years ago

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A mountain climber started at sea level and descended down a cliff. Her location can be represented by -75 feet. How many feet d

id the mountain climber travel?

1 answer:

mariarad [96]3 years ago

8 0

The mountain climber has travelled 75 feet below sea level while descending down a cliff.

Given:

A mountain climber at sea level starts descending down a cliff.
Her location can be represented by -75 ft.

To find:

The distance travelled by a mountain climber.

Solution

Distance ascending above sea level is taken in a positive direction.
Distance descending down sea level is taken in a negative direction.
This negative and positive only represents the direction

The location of mountain climber after descending from cliff = -75 ft

The distance covered by the mountain climber = 75 ft

The mountain climber has travelled 75 feet below sea level while descending down a cliff.

Learn more about descending and ascending here:

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What is the length of leg y of the right triangle?

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Answer:

13

Step-by-step explanation:

We need to use the Pythagorean Theorem to solve this problem:

$a^{2} +b^{2} = c^{2}$ , where a and b are the legs of the right triangle and c is the hypotenuse

In this case, we already know b (b = 84) and c (c = 85), so we need to solve for a:

$a^{2} = c^{2} - b^{2} = 85^{2} - 84^{2} = 169 = 13^{2}$

Thus, a = y = 13.

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3 years ago

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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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lana66690 [7]

Answer:

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Angle 3 = 70 degrees

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Step-by-step explanation:

<u>Finding angle 2</u>

In the figure attached below,the angles 1 and 2 Corresponding angles

When two lines are crossed by another line that is called the Transversal, then, the angles in matching corners are called corresponding angles.

Corresponding angles are equal so

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<u>Finding angle 3</u>

Similarly 2 and 3 are consecutive angles and Consecutive angles are supplementary . That is the sum of the consecutive angles is 180 degree

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angle 3 = 180 -110

angle 3 = 70 degrees

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