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2 years ago

Arlene sleeps for

Mathematics

2 answers:

True [87]2 years ago

7 0

Answer:

50.40

Step-by-step explanation:

7 hours and 20 minutes times 7= 50 hours and 40 minutes

And then you have to change it into a mixed number and it is 50.40

frez [133]2 years ago

6 0

$51 \frac{ 1}{3}$

This equals 51 hours and 20 minutes

You might be interested in

On a cm grid, point p has coordinates (3,-1) and point q has coordinates (-5,6) calculate the shortest distance between p and q

Effectus [21]

Answer:

≈ 10.6 units

Step-by-step explanation:

Calculate the distance d using the distance formula

d = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2 }$

with (x₁, y₁ ) = p(3, - 1) and (x₂, y₂ ) = q(- 5, 6)

d = $\sqrt{(-5-3)^2+(6+1)^2}$

= $\sqrt{(-8)^2+ 7^2}$

= $\sqrt{64+49}$

= $\sqrt{113}$ ≈ 10.6 ( to 1 dec. place )

8 0

2 years ago

Chad owns twice as many cds as carlin owns. if you add 6 to the number of cds chad owns and then divide by 7, you get the number

Luba_88 [7]

35 cds does carlin have

5 0

2 years ago

(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

2 years ago

Devon purchased a new car valued at $16,000 that depreciated continuously at a rate of 35%. Its current value is $2,000. The equ

77julia77 [94]

Answer:

Step-by-step explanation:

Use the equation 2,000 = 16,000(1-r)^t to solve for t;

2000 = 16000(1-0.35)^t

Divide both sides by 16000

2000/16000 = 0.65^t

0.125 =0.65^t

Introduce logarithm on both sides;

<em>ln</em> 0.125 = t <em>ln</em> 0.65

Divide both sides by <em>ln</em> 0.65;

(<em>ln</em> 0.125) / (<em>ln</em> 0.65) = t

-2.07944/ -0.4308 = t

4.827 = t

t= 5 (as a whole number)

Therefore, the car is about 5 years old.

5 0

2 years ago

Jerry and Katja are running in a race. Jerry runs really fast for a while. Then, he stops to catch his breath. After he has rest

Elena L [17]

The slope of the lines represent the runners' speeds. Therefore, the purple line represents Jerry, since he started off faster but also stopped in the middle, and the slope became 0 (horizontal portion of the line.)
The red one represents Katja because she started and ended at a constant pace, which is visible in the red line's constant slope.

4 0

2 years ago

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