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inysia [295]
3 years ago
9

I need help with this one I’m not sure if I got it right

Mathematics
1 answer:
weqwewe [10]3 years ago
7 0

Answer:

144 feet

Step-by-step explanation:

Since this is a quadratic equation, we simply need to find the vertex in order to determine the maximum height!

To solve the equation, follow these steps!:

0 = 16t^2+96t\\\\\frac{-b}{2a} =x\\\\\frac{-96}{2(-16)}=3=x

Now we insert x back into the equation...

= -16(3)^2+96(3)\\ = -16(9)+288\\ =-144+288\\=144

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At what values of x does f(x) = x^3 - 2x^2 -4x+1 satisfy the mean value theorwm on [0,1]
denis-greek [22]

Answer:

x=1/3

Step-by-step explanation:

A function f is given as

f(x) = x^3-2x^2-4x+1 in the interval  [0,1]

This function f being an algebraic polynomial is continuous in the interval [0,1] and also f is differntiable in the open interval (0,1)

Hence mean value theorem applies for f in the given interval

f(1) = 1-2-4+1 = -4\\f(0) = 1

The value

\frac{f(1)-f(0)}{1-0} =\frac{-4-1}{1} =-5

Find derivative for f

f'(x) = 3x^2-4x-4

Equate this to -5 to check mean value theorem

3x^2-4x-4=-5\\3x^2-4x+1=0\\\\(x-1)(3x-1) =0\\x= 1/3 : x = 1

We find that 1/3 lies inside the interval (0,1)

4 0
3 years ago
The price of patatoes first increased by 10% and then fell by 8%. Find the net percentage change in the price of patatoes.
timofeeve [1]

Answer:

10%-8%=2

2/8=0.25

0.25×100=25

The net percentage is 25%

6 0
3 years ago
A circle is graphed on a coordinate grid. Its center is at (3, 4), and the circle passes through point (3, 1). What is the appro
noname [10]

Answer:

i have the same question

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
If the width of the rectangle is three times the length and the perimeter of the rectangle is 72 ft, what are the length and wid
goblinko [34]
W=3L
W+W+L+L=72
Replace W with 3L: 3L+3L+L+L=72
8L=72
L=9
W=27
8 0
3 years ago
How would you determine the axis of symmetry in order to graph x^2 = 8y^2.
Tanya [424]
The picture illustrates the definition. The point P is a typical point on the parabola so that its distance from the directrix, PQ, is equal to its distance from F, PF. The point marked V is special. It is on the perpendicular line from F to the directix. This line is called the axis of symmetry of the parabola or simply the axis of the parabola and the point V is called the vertex of the parabola. The vertex is the point on the parabola closest to the directrix.

Finding the equation of a parabola is quite difficult but under certain cicumstances we may easily find an equation. Let's place the focus and vertex along the y axis with the vertex at the origin. Suppose the focus is at (0,p). Then the directrix, being perpendicular to the axis, is a horizontal line and it must be p units away from V. The directrix then is the line y=-p. Consider a point P with coordinates (x,y) on the parabola and let Q be the point on the directrix such that the line through PQ is perpendicular to the directrix. The distance PF is equal to the distance PQ. Rather than use the distance formula (which involves square roots) we use the square of the distance formula since it is also true that PF2 = PQ2. We get

<span>(x-0)2+(y-p)2 = (y+p)2+(x-x)2. 
x2+(y-p)2 = (y+p)2.</span>If we expand all the terms and simplify, we obtain<span>x2 = 4py.</span>

Although we implied that p was positive in deriving the formula, things work exactly the same if p were negative. That is if the focus lies on the negative y axis and the directrix lies above the x axis the equation of the parabola is

<span>x2 = 4py.</span><span>The graph of the parabola would be the reflection, across the </span>x<span> axis of the parabola in the picture above. A way to describe this is if p > 0, the parabola "opens up" and if p < 0 the parabola "opens down".</span>

Another situation in which it is easy to find the equation of a parabola is when we place the focus on the x axis, the vertex at the origin and the directrix a vertical line parallel to the y axis. In this case, the equation of the parabola comes out to be

<span>y2 = 4px</span><span>where the directrix is the verical line </span>x=-p and the focus is at (p,0). If p > 0, the parabola "opens to the right" and if p < 0 the parabola "opens to the left". The equations we have just established are known as the standard equations of a parabola. A standard equation always implies the vertex is at the origin and the focus is on one of the axes. We refer to such a parabola as a parabola in standard position.

Parabolas in standard positionIn this demonstration we show how changing the value of p changes the shape of the parabola. We also show the focus and the directrix. Initially, we have put the focus on the y axis. You can select on which axis the focus should lie. Also you may select positive or negative values of p. Initially, the values of p are positive. 

5 0
4 years ago
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