All categories

2 years ago

Evaluate the function Find f(2) f(2) =

Mathematics

1 answer:

earnstyle [38]2 years ago

7 0

F(2) is the x value, so you need to look at where the line is in the y direction at x = 2

The black dot on the graph is at x = 2 and is at y = 4

Answer: f(2) = 4

You might be interested in

How would I find the equation of the line of best fit for this table x 7 9 11 y 12.4 13 16

Dominik [7]

Answer:

Use your calculator to find the line of best fit for the data. ... Age Range, \begin {align*}x\end{align*}, 1-3, 4-6, 7-10, 11-14, 15- ... Here is a sample table for the scatterplot

4 0

3 years ago

Given that angle a = 72° and angle b = 37°, work out x.

Sedbober [7]

Answer:

x is 71

Step-by-step explanation:

X + 72 + 37 = 180 ( Sum of angles in a triangle)

X = 180 - 72 - 37

X = 71

8 0

1 year ago

Determine whether the polygons are similar. Similar Not Similar

kupik [55]

The answer is Not similar

6 0

2 years ago

Read 2 more answers

ILL GIVE BRAINLIEST... HURRY PLZ

Alexeev081 [22]

12 Pairs

4 pairs = 8

2 pairs = 4

8 + 4 = 12

If you need total socks, there's 24 total

7 0

2 years ago

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

3 years ago

Read 2 more answers