Combine like radicals to get your answer. image attached

Mathematics

1 answer:

Lena [83]3 years ago

3 0

I believe it’s -6 square root of 2 + 3 square root of 4 + 6 square root of 5

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A function h(x) is defined by the formula h(x)=x^3-3. What is the value of g(h(2))?

Sauron [17]

Answer:

g(h(2))= 5

Step-by-step explanation:

h(x)=x^3-3

g(2^3-3)

the 2 replaces the x

simplify

g(5)

7 0

3 years ago

Let us have four distinct collinear points $a,$ $b,$ $c,$ and $d$ on the cartesian plane. the point $c$ is such that $\dfrac{ab}

Deffense [45]

Start with a line segment connecting two points, A and B. $\dfrac{DA}{BA}=3$ means DA is 3 times longer than BA. Clearly, D cannot fall between A and B because that would mean DA is shorter than BA. So there are two possible locations where D can be placed on the line relative to A and B.

But with $\dfrac{DB}{BA}=2$ , or the fact that DB is 2 times longer than BA, we can rule out one of these positions; referring to the attachment, if we place D to the left of A, then DB would be 4 times longer than BA.

Finally, $\dfrac{AB}{CB}=\dfrac12$ , so that CB is 2 times longer than AB. Again we have two possible locations for point C (it cannot fall between A and B), but one of them forces C to occupy the same point as D. However, A, B, C, D are distinct, so C must fall to the left of A.

Now let $d$ be the length of AB. Then the length of CD in terms of $d$ is $4d$ . We have the coordinates of C and D, and the distance between them is $\sqrt{(4-0)^2+(0-4)^2}=4\sqrt2$ . So

$4d=4\sqrt2\implies d=\sqrt2$

The slope of the line through C and D is

$\dfrac{0-4}{4-0}=-1$

and so the equation of the line through these points is

$y-4=-(x-0)\implies x+y=4$

So the coordinates of A are $(x,y)=(x,4-x)$ . The distance between C and A is $d=\sqrt2$ , so we have

$\sqrt{(x-0)^2+(4-x-4)^2}=\sqrt{2x^2}=|x|\sqrt2=\sqrt2\implies|x|=1$

Since A falls to the right of C (in the $x,y$ plane, not just in the sketch), we know to take the positive value $x=1$ . Then the $y$ coordinate is $y=4-1=3$ .