1:12 p.m to 8:11 p.m

What is the second step in sketching the graph of a rational function?

Elodia [21]

Answer:

C. Choosing test numbers to the left and right of each of the function's zeros and finding the value of the function at each test number

Step-by-step explanation:

Rational function: In mathematics, the term "rational function" is determined as any specific function that can be described through any "rational fraction", in other words, an algebraic fraction, involving both the denominator and the numerator are considered as polynomials. Therefore, the coefficients of specific polynomials are ought not to be in rational numbers and can be considered in the K field.

In the question above, the given statement represents option C is correct.

3 0

3 years ago

Please helppp meeeeee

alexira [117]

Answer: 16) $\bold{\sqrt{x}}$ (17) $\bold{\dfrac{x-4}{2}}$ (18) $\bold{\sqrt{x}-2}$ (19) 3x - 18

<u>Step-by-step explanation:</u>

Inverse is when you swap the x's and y's and then solve for y

16) Given: y = x²

Swapped: x = y²

$\bold{\sqrt{x} =y}$

17) Given: y = 2x + 4

Swapped: x = 2y + 4

x - 4 = 2y

$\bold{\dfrac{x-4}{2}=y}$

18) Given: y = (x + 2)²

Swapped: x = (y + 2)²

√x = y + 2

$\bold{\sqrt{x}-2=y}$

19) Given: $y=\dfrac{1}{3}x+6$

Swapped: $x=\dfrac{1}{3}y+6$

$x - 6 = \dfrac{1}{3}y$

3(x - 6) = y

3x - 18 = y

4 0

2 years ago

Verifying Properties of logarithms In Exercise,use a graphing utility to verify that the functions are equivalent for x > 0.

Ede4ka [16]

Answer:

Applying Quotient and power rule. Graphically, for x>0 both curves coincide.

Step-by-step explanation:

1) Firstly, by applying the Quotient and then the Power Rule, we have a subtraction of the argument and, finally the exponent turns to be the coefficient. As it follows:

$lnx=log_{e}x\\\\f(x) = ln({\frac{x^{2}}{4}})\Rightarrow f(x)=ln{x^2}-ln4\Rightarrow f(x)=2lnx-ln4\\g(x) = 2 ln x - ln 4$

2) Check the graph below to see this equivalence. to see x>0 they both coincide.

7 0

2 years ago

A large truck has two fuel tanks, each with a capacity of 150 gallons. Tank 1 is half full, and Tank 2 is empty.

12345 [234]

Answer:

1) $G(t) = 75 + \frac{23}{4}t\\$

2) After 8 minutes delivering fuel the tanks will have 121 gallons of fuel

Step-by-step explanation:

1) For generating the equation we have to take into account that in the tanks there is a initial volume of fuel that corresponds to 75 gallons, as it is stated that tank one is half full. As the capacity for tank 1 is of 150 gallons, half of the tank equals to:

$\frac{150}{2} = 75 gallons$

Now we have to convert the rate of delivery that is expressed as a mixed number to an improper fraction so:

$5\frac{3}{4} = \frac{(5x4)+3}{4} = \frac{23}{4}$

Then the pumping rate is of 23/4 gallons per minute, to get how many gallons are in the tank we just need to multiply this rate by the time in minutes, and as there is an initial volume we have to add it, so we have the following equation:

$G(t) = 75 + \frac{23}{4}t\\$

2) To know how much fuel is in the tank after 8 minutes we have to replace this time in the previous equation so we have

$G(t) = 75 + \frac{23}{4}t\\G(8) = 75 + \frac{23}{4}(8)\\G(8) = 75 + 23(2)\\G(8) = 75 + 46\\G(8) = 121 gallons$

After 8 minutes delivering fuel the tanks will have 121 gallons of fuel

8 0

2 years ago

Evaluate each expression <br> 8-(-5)-3

Vesnalui [34]

<h3>Answer:</h3>

$\large\boxed{10}$

<h3>Step-by-step explanation:</h3>

In this question, we're evaluating the given expression.

The given expression is:

$8-(-5)-3$

We would simply solve the equation until we reach our final answer.

Solve:

$8-(-5)-3\\\\\text{The "-(-" will turn to a plus sign}\,\,\\\\8+5-3\\\\\text{Add 8 and 5}\\\\13-3\\\\\text{Finish off by subtracting}\\\\10$

A friendly reminder is when you see something like "-(-", you would turn it into a plus sign.

When you're done solving, you should get 10.

This means that 10 would be your final answer.

<h3>I hope this helped you out.</h3><h3>Good luck on your academics.</h3><h3>Have a fantastic day!</h3>

5 0

3 years ago

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