All categories

3 years ago

What is the difference between linear and exponential function?

1 answer:

5 0

Linear functions change at a constant rate per unit interval and take the form y=mx+b, while exponential functions change by a common ratio over equal intervals and take the form y=a^x.

You might be interested in

The first step when dividing 9m2 − 4m − 6 by 3m is shown. Which could be the next step? 3m − − 3m − − 3m − − 3m − −

Agata [3.3K]

Given:

Consider the below picture attached with this question.

The given expression is:

$\dfrac{9m^2}{3m}-\dfrac{4m}{3m}-\dfrac{6}{3m}$

To find:

The next step.

Solution:

We have,

$\dfrac{9m^2}{3m}-\dfrac{4m}{3m}-\dfrac{6}{3m}$

Cancel out the common factors.

$=3m-\dfrac{4}{3}-\dfrac{2}{m}$

Therefore, the correct option is D.

8 0

3 years ago

What set of transformations could be applied to rectangle ABCD to create A″B″C″D″?

morpeh [17]

Answer:

C) Reflected over the x-axis and rotated 90° counterclockwise

Step-by-step explanation:

7 0

3 years ago

The area of a rectangle is 96 in2. The length of the rectangle is 8 inches. Find the perimeter of the rectangle. Show all work f

sukhopar [10]

Length.
96÷8=12inches

Perimeter.
(12+8)+(12+8)
20+20=40inches

8 0

3 years ago

Use the function below to find f(2). f(x)=1/3•4^x

xenn [34]

To find f(2), you just need to replace x for 2.
1/3•4^(2)
The answer is 16/3 or 5.3333...

6 0

3 years ago

Read 2 more answers

A wire b units long is cut into two pieces. One piece is bent into an equilateral triangle and the other is bent into a circle.

mezya [45]

1. Divide wire b in parts x and b-x.

2. Bend the b-x piece to form a triangle with side (b-x)/3

There are many ways to find the area of the equilateral triangle. One is by the formula A= $\frac{1}{2}sin60^{o}side*side= \frac{1}{2} \frac{ \sqrt{3} }{2} (\frac{b-x}{3}) ^{2}= \frac{ \sqrt{3} }{36}(b-x)^{2}$
A= $\frac{ \sqrt{3} }{36}(b-x)^{2}=\frac{ \sqrt{3} }{36}( b^{2}-2bx+ x^{2} )=\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}$

Another way is apply the formula A=1/2*base*altitude,
where the altitude can be found by applying the pythagorean theorem on the triangle with hypothenuse (b-x)/3 and side (b-x)/6

3. Let x be the circumference of the circle.

$2 \pi r=x$

so $r= \frac{x}{2 \pi }$

Area of circle = $\pi r^{2}= \pi ( \frac{x}{2 \pi } )^{2} = \frac{ \pi }{ 4 \pi ^{2} }* x^{2} = \frac{1}{4 \pi } x^{2}$

4. Let f(x)= $\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}+\frac{1}{4 \pi } x^{2}$

be the function of the sum of the areas of the triangle and circle.

5. f(x) is a minimum means f'(x)=0

f'(x)= $\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x$ =0

$\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x=0$

$(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) x=\frac{ \sqrt{3} }{18}b$

$x= \frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

6. So one part is $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$ and the other part is b- $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

4 0

4 years ago

Read 2 more answers