Here is my answer to ur question

F(x)=3x^3-4x^2+6x and g(x)=5x^3+2x^2-3x what is f(X) AND G (X)

kvasek [131]

This on is very easy go to my website so i can explain it better for you

https://www.youtube.com/channel/UCaxhPMTZBmFcxAZ4hYpBf_w

Hope this

8 0

3 years ago

What is the height of cone S?

balu736 [363]

Answer:

12 cm

Step-by-step explanation:

First, we find the scale factor from cone S to cone T.

ratio of volumes = (vol of T)/(vol of S) = (6144 pi cm^3)/(768 pi cm^3) = 8

The ratio of the volumes is 8:1

The scale factor, which is the ratio of linear dimensions (height, radius, etc.), is the cubic root of the ratio of the volumes.

scale factor = cubic root of 8 = 2

The height of cube T is 24 cm, so the height of cube S is 24 cm/2 = 12 cm.

4 0

3 years ago

Read 2 more answers

I’m supposed to find the slope intercept form. The x intercept is (2,0) and the y intercept is (0,6). I keep getting 6/2. But, t

Lilit [14]

$\bf \stackrel{x-intercept}{(\stackrel{x_1}{2}~,~\stackrel{y_1}{0})}\qquad \stackrel{y-intercept}{(\stackrel{x_2}{0}~,~\stackrel{y_2}{6})} \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{6-0}{0-2}\implies \cfrac{6}{-2}\implies -3 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-0=-3(x-2)\implies y=-3x+6$

5 0

3 years ago

Kojo wants to be an author. To improve his writing skills, every day after school he writes 3 pages of a story he is

koban [17]

If the number of days is d and he writes 3 pages for every d, then we can say

3d = 81

3 0

3 years ago

If vector u has its initial point at (-7, 3) and its terminal point at (5, -6), u =

attashe74 [19]

First of all, let θθ be some angle in (0,π)(0,π). Then

θθ is acute ⟺⟺ θ<π2θ<π2 ⟺⟺ cosθ>0cos⁡θ>0.θθ is right ⟺⟺ θ=π2θ=π2 ⟺⟺ cosθ=0cos⁡θ=0.θθ is obtuse ⟺⟺ θ>π2θ>π2 ⟺⟺ cosθ<0cos⁡θ<0.

Now, to see if (say) angle AA of the triangle ABCABC is acute/right/obtuse, we need to check whether cos∠BACcos⁡∠BAC is positive/zero/negative. But what is cos∠BACcos⁡∠BAC? It is the angle made by the vectors AB−→−AB→ and AC−→−AC→. (When you are computing the angle at a particular vertex vv, you should make sure that both the vectors corresponding to the two adjacent sides have that vertex vv as the initial point.) We will first compute these two vectors:

AB−→−=(0,0,0)−(1,2,0)=(−1,−2,0)AB→=(0,0,0)−(1,2,0)=(−1,−2,0)AC−→−=(−2,1,0)−(1,2,0)=(−3,−1,0)AC→=(−2,1,0)−(1,2,0)=(−3,−1,0)Therefore, the angle between these vectors is given by:cos∠BAC=AB−→−⋅AC−→−|AB−→−||AC−→−|=…(1)(1)cos⁡∠BAC=AB→⋅AC→|AB→||AC→|=…Can you take it from here? From the sign of this value, you should be able to decide if angle AA is acute/right/obtuse.

Now, do the same procedure for the remaining two angles BB and CC as well. That should help you solve the problem.

A shortcut. Since you are not interested in the actual values of the angles, but you need only whether they are acute, obtuse or right, it is enough to compute only the sign of the numerator (the dot product between the vectors) in formula (1). The denominator is always positive.

6 0

3 years ago