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Rufina [12.5K]
2 years ago
12

In a volatile housing market, the overall value of a home can be modeled by V(x)=325x2-4600x+145000, where V represents the valu

e of the home and x
represents each year after 2020. Find the vertex and interpret what the vertex of this function means in terms of the value of the home. Show the work you completed
to determine the vertex
Mathematics
1 answer:
shtirl [24]2 years ago
4 0

Answer:

The vertex of the given equation is    which shows that the value of the market after about 5 years is about 187,253.01

Functions and values

Given the overall value of a home can be modeled by

V(x) = 415x^2 – 4600x + 200000

Write in vertex form to have:

V(x) = 415x^2 – 4600x + 200000

For the given equation, the vertex occur at

Hence the vertex of the given equation is    which shows that the value of the market after about 5 years is about 187,253.01

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6x-4y=14 <br> 2x+8y=21 <br> Solve by elimination
Viktor [21]
I wrote some notes please read and if you not sure contact me

8 0
2 years ago
the estimated probability of a football players scoring a touchdown during a particular game is 26% is several simulations of a
joja [24]
Given
   26% = several simulations of a football player playing two games 
Find the percentage of the simulations in each of the two games

2x = .26

2x = .26
----    -----
 2       2

x = .13 * 100
x= 13 

The answer is 13% of simulations would the football player be most likely to get a touchdown in each of the two games. 


4 0
3 years ago
The coordinates of a particle in the metric xy-plane are differentiable functions of time t with:
frez [133]

Answer:

The rate of change of the distance \frac{ds}{dt} when x = 9 and y = 12 is -7.6\:\frac{m}{s}.

Step-by-step explanation:

This is an example of a related rate problem. A related rate problem is a problem in which we know one of the rates of change at a given instant \frac{dx}{dt} and we want to find the other rate \frac{dy}{dt} at that instant.

We know the rate of change of x-coordinate and y-coordinate:

\frac{dx}{dt}=-2\:\frac{m}{s} \\\\\frac{dy}{dt}=-8\:\frac{m}{s}

We want to find the rate of change of the distance \frac{ds}{dt} when x = 9 and y = 12.

The distance of a point (x, y) and the origin is calculated by:

s=\sqrt{x^2+y^2}

We need to use the concept of implicit differentiation, we differentiate each side of an equation with two variables by treating one of the variables as a function of the other.

If we apply implicit differentiation in the formula of the distance we get

s=\sqrt{x^2+y^2}\\\\s^2=x^2+y^2\\\\2s\frac{ds}{dt}= 2x\frac{dx}{dt}+2y\frac{dy}{dt}\\\\\frac{ds}{dt}=\frac{1}{s}(x\frac{dx}{dt}+y\frac{dy}{dt})

Substituting the values we know into the above formula

s=\sqrt{9^2+12^2}=15

\frac{ds}{dt}=\frac{1}{15}(9(-2)+12(-8))\\\\\frac{ds}{dt}=\frac{1}{15}\left(-18-96\right)\\\\\frac{ds}{dt}=\frac{1}{15}(-114)=-\frac{38}{5}=-7.6\:\frac{m}{s}

The rate of change of the distance \frac{ds}{dt} when x = 9 and y = 12 is -7.6\:\frac{m}{s}

7 0
3 years ago
Which is the equation of a parabola with vertex (0, 0), that opens to the left and has a focal width of 12?
sattari [20]

Answer:

\text{Equation of parabola is }y^2=-12x

Step-by-step explanation:

We need to find the equation of parabola using given information

  • Vertex: (0,0)
  • Open to the left
  • Focal width = 12

If parabola open left and passes through origin then equation is

y^2=-4ax

Focal width = 12

Focal width passes through focus and focus is mid point of focal width.

Focus of above parabola would be (-a,0)

Passing point on parabola (-a,6) and (-a,-6)

Now we put passing point into equation and solve for a

6^2=-4a(-a)

a=\pm 3

a can't be negative.

Therefore, a=3

Focus: (-3,0)

Equation of parabola:

y^2=-12x

Please see the attachment of parabola.

\text{Thus, Equation of parabola is }y^2=-12x

6 0
3 years ago
Read 2 more answers
What is the equation of the line that passes through the point (5,6) and has a slope of 2?
pantera1 [17]

Answer:

Answer is at the bottom!!

Step-by-step explanation:

y = - \frac{2}{5} x - 2

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y-intercept )

rearrange 2x + 5y = 10 into this form

subtract 2x from both sides  

5y = - 2x + 10 ( divide all terms by 5 )

y = - \frac{2}{5} x + 2 ← point- slope form with slope m = - \frac{2}{5}

Parallel lines have equal slopes hence  

y = - \frac{2}{5} x + c is the partial equation of the parallel line

to find c, substitute ( 5, - 4 ) into the partial equation

- 4 = - 2 + c ⇒ c = - 4 + 2 = - 2

y = - \frac{2}{5} x - 2 ← equation of parallel line

Hope this helps!!

6 0
3 years ago
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