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3 years ago

Just trying to get through this class from hell

Mathematics

1 answer:

leva [86]3 years ago

8 0

g(x)=4x^2-3x+7

g(3)=4(3)^2-3(3)+7

(4)(3)(3)-3(3)+7

4*9-9+7

36-2

g(3)= 34

Answer: g(3)=34

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Suppose f(x) = x. Find the graph of f(x+2).

Ostrovityanka [42]

<h2>Hello!</h2>

The answer is:

The fourth graph.

We are given the following equation:

$f(x)=x$

Then, we need to find the graph of f(x+2) that will be equal to:

$f(x+2)=x+2$

$y=x+2$

Now, finding the x-axis interception, making "y" equal to 0, we have:

$0=x+2$

$-2=x$

Therefore, the x-axis interception point is located at (-2,0), it matches with the fourth graph.

Now, finding the y-axis interception, making "x" equal to 0, we have:

$y=x+2$

$y=0+2=2$

Therefore, the y-axis interception point is located at (0,2), it matches with the fourth graph.

Hence, we have that the graph of f(x+2) is the fourth graph since the second graph function cuts the x-axis at -2 and the y-axis at 2.

Have a nice day!

7 0

2 years ago

Two cars traveled equal distances in different amounts of time. Car A traveled the distance in 4 h, and Car B traveled the dista

sweet [91]

Sorry I know this late, but the answer is 40 MPH, I don't know if this helps! I HOPE SO THO!! <3

3 0

3 years ago

The center of a circle lies on the line y = 4x + 1 and is tangent to the x-axis at (−3,0) . What is the equation of the circle i

Hunter-Best [27]

Answer:

$(x+3)^2+(y+11)^2=121$

Step-by-step explanation:

Consider a sketch of the problem as shown in the picture, where:

Blue line is given by y = 4x + 1.
Point B is the center of the circle.
Point A is (-3, 0).

Since the center of the circle lies on the line y = 4x +1 and is tangent to the x-axis at point A, then its radius BA is perpendicular to the x-axis. To find the coordinates of point B, we must replace x = -3 into the blue line equation: y = 4x(-3) + 1 = -11.

So, we know that the center of the circle is at B=(-3, -11). And furthermore, the radius BA is of length r=11.

Since the <em>general equation of the circle</em> of radius lenght r centered at (h, k) is given by

$(x-h)^2+(y-k)^2=r^2$