All categories

1 year ago

Scenario #1: Raul

Mathematics

1 answer:

Lorico [155]1 year ago

3 0

Answer:

Compound Interest Answer: £704

You might be interested in

(i)Express 2x² – 4x + 1 in the form a(x+ b)² + c and hence state the coordinates of the minimum point, A, on the curve y= 2x² 4x

earnstyle [38]

Answer:

(i). $y = 2\, x^2 - 4\, x + 1 = 2\, (x - 1)^2 - 1$ . Point $A$ is at $(1, \, -1)$ .

(ii). Point $Q$ is at $\displaystyle \left(-\frac{1}{2},\, \frac{7}{2}\right)$ .

(iii). $\displaystyle y= - \frac{1}{5}\, x + \frac{17}{5}$ (slope-intercept form) or equivalently $x + 5\, y - 17 = 0$ (standard form.)

Step-by-step explanation:

<h3>Coordinates of the Extrema</h3>

Note, that when $a(x + b)^2 + c$ is expanded, the expression would become $a\, x^2 + 2\, a\, b\, x + a\, b^2 + c$ .

Compare this expression to the original $2\, x^2 - 4\, x + 1$ . In particular, try to match the coefficients of the $x^2$ terms and the $x$ terms, as well as the constant terms.

For the $x^2$ coefficients: $a = 2$ .
For the $x$ coefficients: $2\, a\, b = - 4$ . Since $a = 2$ , solving for $b$ gives $b = -1$ .
For the constant terms: $a \, b^2 + c = 1$ . Since $a = 2$ and $b = -1$ , solving for $c$ gives $c =-1$ .

Hence, the original expression for the parabola is equivalent to $y = 2\, (x - 1)^2 - 1$ .

For a parabola in the vertex form $y = a\, (x + b)^2 + c$ , the vertex (which, depending on $a$ , can either be a minimum or a maximum,) would be $(-b,\, c)$ . For this parabola, that point would be $(1,\, -1)$ .

<h3>Coordinates of the Two Intersections</h3>

Assume $(m,\, n)$ is an intersection of the graphs of the two functions $y = 2\, x^2- 4\, x + 1$ and $x -y + 4 = 0$ . Setting $x$ to $m$ , and $y$ to $n$ should make sure that both equations still hold. That is:

$\displaystyle \left\lbrace \begin{aligned}& n = 2\, m^2 - 4\, m + 1 \\ & m - n + 4 = 0\end{aligned}\right.$ .

Take the sum of these two equations to eliminate the variable $n$ :

$n + (m - n + 4) = 2\, m^2 - 4\, m + 1$ .

Simplify and solve for $m$ :

$2\, m^2 - 5\, m -3 = 0$ .

$(2\, m + 1)\, (m - 3) = 0$ .

There are two possible solutions: $m = -1/2$ and $m = 3$ . For each possible $m$ , substitute back to either of the two equations to find the value of $n$ .

$\displaystyle m = -\frac{1}{2}$ corresponds to $n = \displaystyle \frac{7}{2}$ .
$m = 3$ corresponds to $n = 7$ .

Hence, the two intersections are at $\displaystyle \left(-\frac{1}{2},\, \frac{7}{2}\right)$ and $(3,\, 7)$ , respectively.

<h3>Line Joining Point Q and the Midpoint of Segment AP</h3>

The coordinates of point $A$ and point $P$ each have two components.

For point $A$ , the $x$ -component is $1$ while the $y$ -component is $(-1)$ .
For point $P$ , the $x$ -component is $3$ while the $y$ -component is $7$ .

Let $M$ denote the midpoint of segment $AP$ . The $x$ -component of point $M$ would be $(1 + 3) / 2 = 2$ , the average of the $x$ -components of point $A$ and point $P$ .

Similarly, the $y$ -component of point $M$ would be $((-1) + 7) / 2 = 3$ , the average of the $y\!$ -components of point $A$ and point $P$ .

Hence, the midpoint of segment $AP$ would be at $(2,\, 3)$ .

The slope of the line joining $\displaystyle \left(-\frac{1}{2},\, \frac{7}{2}\right)$ (the coordinates of point $Q$ ) and $(2,\, 3)$ (the midpoint of segment $AP$ ) would be:

$\displaystyle \frac{\text{Change in $y$}}{\text{Change in $x$}} = \frac{3 - (7/2)}{2 - (-1/2)} = \frac{1}{5}$ .

Point $(2,\, 3)$ (the midpoint of segment $AP$ ) is a point on that line. The point-slope form of this line would be:

$\displaystyle \left( y - \frac{7}{2}\right) = \frac{1}{5}\, \left(x - \frac{1}{2} \right)$ .

Rearrange to obtain the slope-intercept form, as well as the standard form of this line:

$\displaystyle y= - \frac{1}{5}\, x + \frac{17}{5}$ .

$x + 5\, y - 17 = 0$ .

7 0

1 year ago

The size, S, of a tumor (in cubic millimeters) is given by S=2^t, where t is the number of months since the tumor was discovered

nordsb [41]

A) you just do 2^9 which is 512 cubic millimeters
b) using the average rate of change formula you get 56.78 cubic millimeters per month
c) I do not understand at all I am stuck on the same one<span />

4 0

2 years ago

A survey was administered to high school seniors in Anytown. According to the survey results, fewer than 0.5% of the students dr

Feliz [49]

Answer: high test-retest reliability

Step-by-step explanation: this is because the result of the survey was thesame with the previous result, despite the space of time between when the first survey was conducted and when the second survey was conducted. There was know observable difference in result and if conducted in the next 3 months again, it will give same result, this strongly indicate that the survey has high test-retest reliability.

8 0

1 year ago

A music competition on television had five elimination rounds. After each elimination, only half of the contestants were sent to

ikadub [295]

Its c, average rate, im sure of it

6 0

2 years ago

Read 2 more answers

Consider the initial value problem: 2ty′=8y, y(−1)=1. Find the value of the constant C and the exponent r so that y=Ctr is the s

VikaD [51]

The correct question is:

Consider the initial value problem

2ty' = 8y, y(-1) = 1

(a) Find the value of the constant C and the exponent r such that y = Ct^r is the solution of this initial value problem.

b) Determine the largest interval of the form a < t < b on which the existence and uniqueness theorem for first order linear differential equations guarantees the existence of a unique solution.

c) What is the actual interval of existence for the solution obtained in part (a) ?

Step-by-step explanation:

Given the differential equation

2ty' = 8y

a) We need to find the value of the constant C and r, such that y = Ct^r is a solution to the differential equation together with the initial condition y(-1) = 1.

Since Ct^r is a solution to the initial value problem, it means that y = Ct^r satisfies the said problem. That is

2tdy/dt - 8y = 0

Implies

2td(Ct^r)/dt - 8(Ct^r) = 0

2tCrt^(r - 1) - 8Ct^r = 0

2Crt^r - 8Ct^r = 0

(2r - 8)Ct^r = 0

But Ct^r ≠ 0

=> 2r - 8 = 0 or r = 8/2 = 4

Now, we have r = 4, which implies that

y = Ct^4

Applying the initial condition y(-1) = 1, we put y = 1 when t = -1

1 = C(-1)^4

C = 1

So, y = t^4

b) Let y = F(x,y)................(1)

Suppose F(x, y) is continuous on some region, R = {(x, y) : x_0 − δ < x < x_0 + δ, y_0 −ę < y < y_0 + ę} containing the point (x_0, y_0). Then there exists a number δ1 (possibly smaller than δ) so that a solution y = f(x) to (1) is defined for x_0 − δ1 < x < x_0 + δ1.

Now, suppose that both F(x, y)

and ∂F/∂y are continuous functions defined on a region R. Then there exists a number δ2

(possibly smaller than δ1) so that the solution y = f(x) to (1) is

the unique solution to (1) for x_0 − δ2 < x < x_0 + δ2.

c) Firstly, we write the differential equation 2ty' = 8y in standard form as

y' - (4/t)y = 0

0 is always continuous, but -4/t has discontinuity at t = 0

So, the solution to differential equation exists everywhere, apart from t = 0.

The interval is (-infinity, 0) n (0, infinity)

n - means intersection.

7 0

1 year ago