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3 years ago

If f(x)=-x^2+6x-1 and g(x) =3x^2-4x-1, find (f-g)(x)

Mathematics

2 answers:

asambeis [7]3 years ago

8 0

Answer:

Step-by-step explanation:

Given,

$f(x)=-x^{2} +6x-1\\g(x)=3x^{2} -4x-1$

we know,

(f-g)(x)=f(x)-g(x)

So, here we get

(f-g)(x)

=f(x)-g(x)

= $-x^{2} +6x-1-(3x^{2} -4x-1)$

= $-x^{2} +6x-1-3x^{2} +4x+1$

= $-x^{2}-3x^{2} +6x+4x-1+1$

= $-4x^{2} +10x$

So, the answer is

$-4x^{2} +10x$

Oksanka [162]3 years ago

8 0

Answer:

(f-g)(x) = -4x^2 + 10x - 2

Step-by-step explanation:

Writing f(x) first and then writing g(x) beneath it may make this problem a bit easier to understand:

f(x)= -x^2+6x -1

-g(x) = -3x^2+4x -1

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

(f-g)(x) = -4x^2 + 10x - 2

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3 years ago

A ball is thrown from a height of 38 meters with an initial downward velocity of 3 m/s

ANTONII [103]

<h2>Hello!</h2>

The answer is:

The ball will hit the ground after $t=2.47s$

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So,

$\frac{-b+-\sqrt{b^{2} -4ac} }{2a}$

We are given the function:

$h=-5t^{2}-3t+38$

Where,

$a=-5\\b=-3\\c=38$

Then, substituting it into the quadratic equation, we have:

$\frac{-b+-\sqrt{b^{2} -4ac} }{2a}=\frac{-(-3)+-\sqrt{(-3)^{2} -4*-5*38} }{2*-5}\\\\\frac{-(-3)+-\sqrt{(-3)^{2} -4*-5*38} }{2*-5}=\frac{3+-\sqrt{9+760} }{-10}\\\\\frac{3+-\sqrt{9+760} }{-10}=\frac{3+-\sqrt{769} }{-10}=\frac{3+-27.731}{-10}\\\\t1=\frac{3+27.731}{-10}=-3.073s\\\\t2=\frac{3-27.731}{-10}=2.473s$

Since negative time does not exists, the ball will hit the ground after:

$t=2.473s=2.47s$

Have a nice day!

4 0

3 years ago

Read 2 more answers

Let f(x)=2x^2−8 .

madam [21]

To translate a function up or down you add a constant to the function...2 units downward means you must subtract 2 from the original equation...

g(x)=f(x)-2

g(x)=2x^2-8-2

g(x)=2x^2-10

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3 years ago