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3 years ago

The excluded values of a rational expression are 2 and 5. Which of the following could be this expression?

Mathematics

2 answers:

SOVA2 [1]3 years ago

6 0

Answer: d on eng

Step-by-step explanation:

irina [24]3 years ago

5 0

A value will be excluded from a rational expression if it causes the denominator to be zero as dividing by zero is undefined.

An example that would work for your specific question is:

$y = \frac{(x - 3)(x-1)}{(x-2)(x-5)}$

If you plug in either 2 or 5 into this equation the denominator will be zero causing the expression to undefined there, so the values 2 and 5 are excluded from the domain of the expression.

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Find the volume of this cylinder.

Aleonysh [2.5K]

Cylinder's volume = 1330.5 cm^3

Step-by-step explanation:

radius, r = diameter, d / 2

= 11/2

= 5.5 cm

Volume = π r^2 h

= π × 5.5^2 × 14

= 423.5π

= 1330.5 cm^3

3 0

3 years ago

Read 2 more answers

AC = Round your answer to the nearest hundredth. A 5 35 B C

baherus [9]

Answer:

2.87 = AC

Step-by-step explanation:

Since this is a right triangle we can use trig functions

sin theta = opp / hyp

sin 35 = AC /5

5 sin 35 = AC

2.867882182= AC

To the nearest hundredth

2.87 = AC

8 0

3 years ago

Is 21q-7 equivalent to (3q-p)(7)

Jobisdone [24]

They are equivalent ! both are 14

4 0

3 years ago

ANSWER QUICKLY!!!!!

Paul [167]

Answer:

1. The possible roots are ±1, ±2, ±3, ±6. These are the factors of the trailing constant (6) divided by the factors of the leading coefficient (1). When the leading coefficient is 1, the possible roots are the factors of the constant

2. The polynomial is easier to evaluate when it is written in Horner form:

f(x) = ((x -4)x +1)x +6

To show that 2 is a zero, we want to find f(2):

f(2) = ((2 -4)2 +1)2 +6 = (-4 +1)2 +6 = -6 +6 = 0

f(2) = 0, so 2 is one of the zeros of this function

3. Using synthetic division (attached) or polynomial long division, we can divide the given polynomial by (x-2) to find the remaining factors. This division gives (x^2 -2x -3), which can be factored as (x -3)(x +1), so the three actual roots are ...

x = 2 (from above), x = 3, x = -1 (from our factorization)

4. In factored form, the polynomial can be written ...

f(x) = (x +1)(x -2)(x -3)

The first factor was found from the fact that 2 was given as a zero of the function. For any zero "a", a factor of the polynomial is (x-a).

The remaining factors were found by factoring the quadratic trinomial that resulted from the division of f(x) by x-2. That trinomial is x^2 -2x -3.

There are a number of methods that can be used to factor x^2 -2x -3. Again, the rational root theorem can help. It suggests that ±1 and ±3 are possible roots.

4 0

2 years ago

Find dy/dx if y =x^3+5x+2/x²-1

stiks02 [169]

Differentiate using the Quotient Rule –

$\qquad$ $\pink{\twoheadrightarrow \sf \dfrac{d}{dx} \bigg[\dfrac{f(x)}{g(x)} \bigg]= \dfrac{ g(x)\:\dfrac{d}{dx}\bigg[f(x)\bigg] -f(x)\dfrac{d}{dx}\:\bigg[g(x)\bigg]}{g(x)^2}}\\$

According to the given question, we have –

f(x) = x^3+5x+2
g(x) = x^2-1

Let's solve it!

$\qquad$ $\green{\twoheadrightarrow \bf \dfrac{d}{dx}\bigg[ \dfrac{x^3+5x+2 }{x^2-1}\bigg]} \\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{(x^2-1) \dfrac{d}{dx}(x^3+5x+2) - ( x^3+5x+2) \dfrac{d}{dx}(x^2-1)}{(x^2-1)^2 }\\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{(x^2-1)(3x^2+5) - ( x^3+5x+2) 2x}{(x^2-1)^2 }\\$

$\qquad$ $\pink{\sf \because \dfrac{d}{dx} x^n = nx^{n-1} }\\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-(2x^4+10x^2+4x)}{(x^2-1)^2 }\\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-2x^4-10x^2-4x}{(x^2-1)^2 }\\$

$\qquad$ $\green{\twoheadrightarrow \bf \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}\\$

$\qquad$ $\pink{\therefore \bf{\green{\underline{\underline{\dfrac{d}{dx} \dfrac{x^3+5x+2 }{x^2-1}} = \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}}}}\\\\$

7 0

2 years ago