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alexdok [17]
3 years ago
15

Let f(x) = tan(x) - 2/x. Let g(x) = x^2 + 8. What is f(x)*g(y)?

Mathematics
1 answer:
Tema [17]3 years ago
5 0

Answer:

f(x)\times g(y)=y^2tan(x)+8tan(x)-\frac{2y^2}{x}-\frac{16}{x}

Step-by-step explanation:

We are given that

f(x)=tan(x)-\frac{2}{x}

g(x)=x^2+8

We have to find f(x)\times g(y)

To find the value of f(x)\times g(y) we will multiply f(x) by g(y)

g(y)=y^2+8

Now,

f(x)\times g(y)=(tanx-\frac{2}{x})(y^2+8)

f(x)\times g(y)=tan(x)(y^2+8)-\frac{2}{x}(y^2+8)

f(x)\times g(y)=y^2tan(x)+8tan(x)-\frac{2y^2}{x}-\frac{16}{x}

Hence,

f(x)\times g(y)=y^2tan(x)+8tan(x)-\frac{2y^2}{x}-\frac{16}{x}

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Which equation is y = –6x^2 + 3x + 2 rewritten in vertex form? y = negative 6 (x minus 1) squared + 8 y = negative 6 (x + one-fo
mart [117]

Answer:

y  = -6(x - \frac{1}{2})^2 -\frac{7}{2}

Step-by-step explanation:

Given:

y = -6x^2 + 3x + 2

Required

Rewrite in vertex form

The vertex form of an equation is in form of: y = a(x - h)^2+ k

Solving: y = -6x^2 + 3x + 2

Subtract 2 from both sides

y - 2 = -6x^2 + 3x + 2 - 2

y - 2 = -6x^2 + 3x

Factorize expression on the right hand side by dividing through by the coefficient of x²

y - 2 = -6(x^2 + \frac{3x}{-6})

y - 2 = -6(x^2 - \frac{3x}{6})

y - 2 = -6(x^2 - \frac{x}{2})

Get a perfect square of coefficient of x; then add to both sides

------------------------------------------------------------------------------------

<em>Rough work</em>

The coefficient of x is \frac{-1}{2}

It's square is (\frac{-1}{2})^2 = \frac{1}{4}

Adding inside the bracket of -6(x^2 - \frac{x}{2}) to give: -6(x^2 - \frac{x}{2} + \frac{1}{4})

To balance the equation, the same expression must be added to the other side of the equation;

Equivalent expression is: -6(\frac{1}{4})

------------------------------------------------------------------------------------

The expression becomes

y - 2 -6(\frac{1}{4})= -6(x^2 - \frac{x}{2} + \frac{1}{4})

y - 2 -\frac{6}{4}= -6(x^2 - \frac{x}{2} + \frac{1}{4})

y - 2 -\frac{3}{2}= -6(x^2 - \frac{x}{2} + \frac{1}{4})

Factorize the expression on the right hand side

y - 2 -\frac{3}{2}= -6(x - \frac{1}{2})^2

y - (2 +\frac{3}{2})= -6(x - \frac{1}{2})^2

y - (\frac{4+3}{2})= -6(x - \frac{1}{2})^2

y - (\frac{7}{2})= -6(x - \frac{1}{2})^2

y  +\frac{7}{2} = -6(x - \frac{1}{2})^2

Make y the subject of formula

y  = -6(x - \frac{1}{2})^2 -\frac{7}{2}

<em>Solved</em>

7 0
4 years ago
Read 2 more answers
If -1 and 2 are the two zeros of the polynomial x3-4x2-7x+10 find the third zero
rewona [7]

note that the zeroes should be x = 1 and x = - 2 not x = - 1 and x = 2

the factors are then ( x - 1 ) and (x + 2)

and x³ - 4x² - 7x + 10 = (x - 1 )(x + 2) = x² + x - 2

thus x² + x - 2 is a factor

dividing x³ - 4x² - 7x + 10 by x² + x - 2 gives x - 5

the third zero is x = 5

check : (5)³ -4(5)² - 7(5) + 10 = 125 - 100 - 35 + 10 = 0




5 0
3 years ago
Which statement is true?
defon

Answer:

b. Multiplying decimals less than one is finding a part of a part.

Step-by-step explanation:

A part of a given number implies a selected part of a given number which can be expressed as a fraction of the number. Example, the half of any number is \frac{1}{2}.

The product of a part and another part can be expressed as;

\frac{1}{3} x \frac{2}{3} = \frac{2}{9}

or,

\frac{1}{2} x \frac{1}{2} = \frac{1}{4}

or,

\frac{1}{2} x \frac{2}{3} = \frac{2}{6} = \frac{1}{3}

The three examples shows: a part x a part.

Therefore, it would be observed that multiplying decimals less than one is the same as finding a part of a part.

5 0
3 years ago
Can someone please help me out with this
Sliva [168]

Answer:

hope it helps

4 0
3 years ago
A taxi charges a base fee plus $0.75/km.
Naily [24]

a) The rate of change in this situation is 0.75 dollars/km

b) The equation for the cost of hiring a taxi in terms of length of the

trip is y = 0.75 x + b

c) The initial value represents in this situation is $1.2

Step-by-step explanation:

The given is:

  • A taxi charges a base fee plus $0.75/km.
  • A 10-km trip costs $8.70

∵ A rate of change is a rate that describes how one quantity

  changes in relation to another quantity

∵ The unit cost of a taxi is 0.75 dollars for each 1 kilometer

∴ The rate of change = 0.75 dollars/km

a) The rate of change in this situation is 0.75 dollars/km

Assume that the cost of hiring a taxi is $y for length of a trip x km

and a base fee of $b

∵ The length of the trip = x km

∵ The cost per km = $0.75

∵ The base fee = $b

∵ The cost of hiring a taxi = $y

- Write an equation for the the cost of hiring a taxi

∴ y = 0.75 x + b

b) The equation for the cost of hiring a taxi in terms of length

of the trip is y = 0.75 x + b

∵ The length of the trip is 10 km

∴ x = 10

∵ The cost of the hiring a taxi is $8.70

∴ y = 8.70

- Substitute these values in the equation of part (b)

∵ y = 0.75 x + b

∴ 8.70 = 0.75(10) + b

∴ 8.70 = 7.5 + b

- Subtract 7.5 from both sides

∴ 1.2 = b

∵ b is the base fee

∴ b is the initial value

∴ The initial value = $1.2

c) The initial value represents in this situation is $1.2

Learn more:

You can learn more about word problems in brainly.com/question/3950386

#LearnwithBrainly

3 0
3 years ago
Read 2 more answers
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