Taylor Starts Practicing Her Math Facts At 4:30. She Is Finished Practicing At 7:00.

1. Given f(x)=2x 2 +1 and g(x)=3x-5 find the following

LenKa [72]

The value of the function 1) f -g = 2x² - 3x + 6 and 2) f(g(2)) is 3.

Here the two functions are given, f(x) and g(x).

f(x) = 2x² + 1

g(x) = 3x - 5

We have to find f-g and f(g(2)).

1) f- g

f(x) - g(x)

(2x² + 1) - ( 3x - 5)

2x² + 1 - 3x + 5

2x² - 3x + 6

2) f(g(2))

f(g(x)) = 2(3x-5)² + 1

= 2( 9x² - 30x + 25) + 1

= 18x² - 60x + 50 + 1

= 18x² - 60x + 51

f(g(2)) = 18(2)² - 60(2) + 51

=18× 4 - 120 + 51

= 72 - 120 + 51

= 123 - 120

= 3

Therefore the value of f-g is 2x²- 3x + 6 and the value of f(g(2)) is 3.

To know more about the function refer to the link given below:

brainly.com/question/11624077

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4 0

1 year ago

Find the Indicated Length of WX! Will get brainlist please explain to and show steps!!1

inessss [21]

If XZ=20 then half of that, XY = 10 so if XY is 10 that means YZ is 10, and WX is 10.

4 0

3 years ago

Which of the following expressions can be used to show 5^3/2 is equivalent to radical 5^3

Nataly_w [17]

Answer:

Option (4) is correct.

Step-by-step explanation:

We need to show that $5^{3/2}$ is equivalent to radical $\sqrt{5^3}$ .

We can do it as follows :

$\sqrt{5^3} =(5^3)^{\dfrac{1}{2}}\\\\=5^{\dfrac{3}{2}}$

Hence, the correct option is (4).

5 0

2 years ago

show that the equation 3x2 + 3y2 + 6x-y= 0 represents a circle and find the centre and radius of the circle

Jlenok [28]

Answer:

Center of the circle: $(-1,\frac{1}{6})$

Radius of the circle: $\frac{\sqrt{37} }{6}$

Step-by-step explanation:

Let's start by dividing both sides of the equation by the factor "3" so we simplify our next step of completing squares for x and for y:

$3x^2+3y^2+6x-y=0\\x^2+y^2+2x-\frac{1}{3} y=0$

Now we work on completing the squares for the expression on x and for the expression on y separately, so we group together the terms in "x" and then the terms in "y":

$x^2+y^2+2x-\frac{1}{3} y=0\\( x^2+2x ) + (y^2-\frac{1}{3} y)=0$

Let's find what number we need to add to both sides of the equation to complete the square of the group on the variable "x":

$( x^2+2x ) = 0\\x^2+2x+1=1\\(x+1)^2=1$

So, we need to add "1" to both sides in order to complete the square in "x".

Now let's work on a similar fashion to find what number we need to add on both sides to complete the square for the group on y":

$(y^2-\frac{1}{3} y)=0\\y^2-\frac{1}{3} y+\frac{1}{36} =\frac{1}{36}\\(y-\frac{1}{6} )^2=\frac{1}{36}$

Therefore, we need to add " $\frac{1}{36}$ " to both sides to complete the square for the y-variable.

This means we need to add a total of $1 + \frac{1}{36} = \frac{37}{36}$ to both sides of the initial equation in order to complete the square for both variables:

$x^2+y^2+2x-\frac{1}{3} y=0\\x^2+y^2+2x-\frac{1}{3} y+\frac{37}{36} =\frac{37}{36} \\(x+1)^2+(y-\frac{1}{6} )^2=\frac{37}{36}$

Now recall that the right hand side of this expression for the equation of a circle contains the square of the circle's radius, based on the general form for the equation of a circle of center $(x_0,y_0)$ and radius R: