Someone help me solve

Mathematics

1 answer:

Lady_Fox [76]3 years ago

3 0

First draw sloping up (steep because he’s running)to the right until you hit 20 on the y axis. Next draw down and to the right for 5 yards until 15 on y axis. Next is a flat horizontal line for 3 seconds on x axis. Finally a shallow slope line down (because he is walking) and right until you hit the x axis.

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If g(x)=4x-6 then g^-1(x)=?<br> (1) -4x+6<br> (2)-1/4x+1/6<br> (3)1/4x+3/2<br> (4)1/4x+6

kati45 [8]

Given:

$g(x)=4x-6$

To find:

The function $g^{-1}(x)$ .

Solution:

we have,

$g(x)=4x-6$

Put g(x)=y.

$y=4x-6$

Interchange x and y.

$x=4y-6$

Isolate y.

$x+6=4y$

$\dfrac{x+6}{4}=y$

$y=\dfrac{x+6}{4}$

Substitute $y=g^{-1}(x)$ .

$g^{-1}(x)=\dfrac{x+6}{4}$

$g^{-1}(x)=\dfrac{x}{4}+\dfrac{6}{4}$

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

Therefore, the correct option is (3).

4 0

3 years ago

Read 2 more answers

Describe the number and type of roots for the polynomial (how many real and complex?). x3 + 5x2 – 4x – 2 = 0

zloy xaker [14]

Descartes' rule of signs tells you the one sign change in the coefficients means there's one positive real root. If you change the sign of the odd-power terms, the signs become -++-, so have two sign changes. This means there are 0 or 2 negative real roots.

A graph shows there to be 3 real roots, two of which are negative. Thus there are no complex roots.