All categories

3 years ago

Is the function ƒ(x) = (4)x – 2 an exponential function? If so, identify the base. If not, why not?

2 answers:

8 0

Answer:no, the function may be rewritten so that the constant multiplier is 4 -2, a positive number raised to a negative exponent.

Step-by-step explanation:this is the answer on GRADPOINT

NARA [144]3 years ago

5 0

No it's not an exponential equation in which f(x)is not qualified to an exponential number. no number from 1 - 20 can be used and get subtracted by 2 to give a square number.
eg-
4×2-2≠ square number

You might be interested in

I need help with this

Ede4ka [16]

I Can't See The Problem All The Way

4 0

2 years ago

A Spaceship traveled 93 billion miles. in scientific notation that is how many miles?

Alex

4 0

2 years ago

If a car changes its velocity from 32 km/hr to 54 km/hr in 8.0 seconds, what is its acceleration?

densk [106]

2.8 km/hr/sec, B is the answer

6 0

3 years ago

Read 2 more answers

A baseball pitch is clocked at 93 miles per hour.Convert this rate to feet per second.

Dmitriy789 [7]

(93 mile/hour) x (5,280 feet/mile) x (1 hour / 3,600sec) = <em>136.4 feet/sec</em> .

3 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:

$(x-x_0)^2+(y-y_0)^2=R^2$

So our equation that can be written as:

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

corresponds to a circle centered at $(-1,\frac{1}{6})$ , and with radius $\sqrt{\frac{37}{36}}=\frac{\sqrt{37} }{6}$

5 0

2 years ago