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2 years ago

The dimensions of a gymnastic floor are 40 ft by 40 ft. the performance floor is 1 ft less, or 39 ft, per side.

Mathematics

1 answer:

Mashutka [201]2 years ago

4 0

Answer:

the square root of 3200

Step-by-step explanation:

I'm not sure if this is right, because the format of the question wasn't clear, but it sounded like we have to find the diagonal distance of the room from one corner to the opposite. If the sides are 40ft long on both sides, we can use the Pythagorean Theorem to solve this:

40^2+40^2=c^2

1600+1600=c^2

3200=c^2

c=sqrt(3200)

or about 56.5

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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.

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

4 0

3 years ago

Olsons sub shop offers 8 vegetables toppings , 6 meat toppings, and three bread choices. How many different sandwiches can you o

kap26 [50]

I THINK you just multiply them all together
8*6*3 = 144

5 0

4 years ago

Let f(x) = |x|, ????(x) = 2f(x), and h(x) = 1 / 2 f(x) for any real number x.

SashulF [63]

Answer:

g(x) = 2|x|

h(x) = $\frac{1}{2}$ |x|

Step-by-step explanation:

If you have g(x)=2f(x) and h(x)= $\frac{1}{2}$ f(x), you can rewrite the expressions in terms of |x| as the following:

- For g(x)=2f(x) you have:

g(x) = 2|x|

-For h(x)= $\frac{1}{2}$ f(x), you have:

h(x) = $\frac{1}{2}$ |x|

8 0

3 years ago

How do you add and subtract radicals? If you could please give me the instructions step by step and with examples, that would be

EleoNora [17]

You can add and subtract radicals by ensuring that they have same radical parts.

<h3>What's radical?</h3>

It should be noted that a radical expression is an expression that contains the radical symbol which is ✓.

The necessary steps to add or subtract radicals include:

Identify the radical parts.

Add them together.

If there's unlike radical parts, one will manipulate the parts involved before solving.

Example: Add ✓63 + 5✓7

It should be noted that ✓63 = ✓9 × ✓7 = 3✓7

Therefore, ✓63 + 5✓7 will be:

= 3✓7 + 5✓7

= 8✓7.

Learn more about radical on:

brainly.com/question/738531

#SPJ1

6 0

1 year ago

Parabolas. Please help me

olganol [36]

x² = 16 y
x = 0
(x + 4)² = 2/3 (y - 2)
Gayle identifies that the vertex of the parabola is (3, -1) . The parabola opens right, and the focus is 3 units away from the vertex. The directrix is 6 units from the focus. The focus is the point (6, -1). The directrix of the equation is x = 0.
(y - 6)² = 4p (x - 3)

Step-by-step explanation:

1. First figure

We plot the parabola as given in the attached diagram.

As it is facing upwards, the equation goes as x² = 4py

where, p = 4 (refer the attached diagram)

x² = 4py

x² = 4 (4) y

∴, standard form of parabola is x² = 16 y

2. Second figure

(y + 3)² = 4 (x - 1)

Comparing the given equation with the standard form

(y - k)² = 4p (x - h)

Now from this equation we get to know that

h = 1

p = 1

Directrix is x = (h - p)

So, x = 0

3. Third figure

3x² + 24x - 2y + 52 = 0

3x² + 24x = 2y - 52

3 (x² + 8x) = 2 (y - 26)

(x² + 8x) = 2 (y - 26) / 3

Adding 16 on both sides,

x² + 8x + 16 = 2 (y - 26) / 3 + 16

(x + 4)² = 2/3 y - 52/3 + 16

(x + 4)² = 2/3 y - 4/3

(x + 4)² = 2/3 (y - 2)

4. Fourth figure

(y + 1)² = 12 (x - 3)

Comparing the given equation with the standard form

(y - k)² = 4p (x - h)

Now from this equation we get to know that

k = -1

h = 3

p = 3

Gayle identifies that the vertex of the parabola is (3, -1) . The parabola opens right, and the focus is 3 units away from the vertex. The directrix is 6 units from the focus. The focus is the point (6, -1). The directrix of the equation is x = 0.

5. Fifth figure

Focus = (2, 6)

Directrix is x = 4

Therefore, it follows the standard form

(y - k)² = 4p (x - h)

Directrix is given by x = h-p = 4

Focus is given by (h + p, k) = (2, 6)

Solving for (h - p) = 4, (h + p) = 2

2 - p - p = 4

-2p = 2

p = -1

Hence, h = 3

Therefore, the standard form can be written as

(y - 6)² = 4p (x - 3)

7 0

4 years ago