Ask question

Ask question

All categories

3 years ago

8

What is the graph of the function? F(x) = 1/2x^2

1 answer:

Sergio [31]3 years ago

7 0

F(x) = 1/2x^2 is a variation of the basic graph y = x^2 (the squaring function).

The graph of the squaring function has the shape of a parabola that opens up. Its vertex is at (0,0).

The graph of F(x) = (1/2)x^2 has the same shape, but is compressed vertically by a factor of (1/2). The vertex of F(x) = (1/2)x^2 is still (0,0). That eliminates the first 2 possible answers. Note that F(1) = (1/2)(1)^2 = 1/2. The point (1, 1/2) is on the third graph, but not on the fourth. Thus, the third graph is correct.

You might be interested in

There are 20 children in the cast of a class play, and 8 of the children are boys. of the boys, 4 have a speaking part in the pl

likoan [24]

The probability that a child with a speaking part is chosen randomly would be 2:5.

6 0

3 years ago

Need help plsssss :;):5:$//&-&!.&/).:$83

Katyanochek1 [597]

The answer is b.

Because you multiply (5•12)

3 0

3 years ago

If a sector of a circle has a radius of 15m and an arc length of 25 meters, find the central

ozzi

Answer:

25

Step-by-step explanation:

5 0

3 years ago

What are the real and complex solutions of the polynomial equation? x^4-41x^2=-400. *work needed*

pickupchik [31]

<h3>Answer: x = -5, -4, 4 and 5.</h3><h3>All four zeros are real solutions.</h3>

Step-by-step explanation:

Given the polynomial equation $x^4-41x^2=-400$ .

Adding 400 on both sides to get rid 400 from right side and set 0 on right side, we get

$x^4-41x^2+400=-400+400$ .

$x^4-41x^2+400=0$ .

Factoring by product sum rule.

We need product of 400 and sum upto -41.

We can see that 400 = -25 × -16 = 400 and -25-16 = -41.

Therefore,

$x^4-25x^2-16x^2+400=0$

Making it into two groups, we get

$(x^4-25x^2)+(-16x^2+400)=0$

Factoring out GCF of each group, we get

$x^2(x^2-25)-16(x^2-25)=0$

$(x^2-25)(x^2-16) =0$

Factoring out $(x^2-25)$ and $(x^2-16)$ separately by difference of the squares identity $a^2-b^2=(a-b)(a+b)$ , we get

$(x^2-25) = x^2-5^2= (x-5)(x+5)$ and

$x^2-16 = x^2-4^2 =(x-4)(x+4)$ .

Therefore,

$(x-5)(x+5)(x-4)(x+4) =0$

Applying zero product rule,

x-5=0

x+5=0

x-4=0 and

x+4=0.

Therefore,

<h3>x = -5, -4, 4 and 5.</h3><h3>All four zeros are real solutions.</h3>

6 0

3 years ago

What is this answer?

kvasek [131]

I got S > 2
Because you would multiply both sides by two then subtract by twelve
_

3 0

4 years ago