3. Name the property the equation illustra

The endpoints of AB are A(2, 3) and B(8, 1). The perpendicular bisector of AB is CD, and point C lies on AB. The length of CD is

marysya [2.9K]

Comment
If C lies on AB Then it must be the midpoint of AB because CD is the Perpendicular Bisector of AB

Midpoint
A(2,3) B(8,1)
Midpoint formula = (x1 + x2)/2 + (y1 + y2)/2
Midpoint = (2 + 8)/2 + (3 + 1) / 2
Midpoint = 10/2 + 4/2
Midpoint = (5,2) <<<<< answer

Slope AB
slope = (y2 - y1)/(x2 - x1)
slope = (3 - 1) / (2 - 8)
slope = 2/-6 = -1/3

Slope CD
CD_slope * AB_slope = - 1
CD_slope * -1/3 = -1 multiply both sides by - 3
CD_slope * 1 = - 1 * -3
CD_slope = 3 <<<<< Slope CD Answer

Equation CD
Given the midpoint of AB which is C( and the slope of CD
y = 3x + b
2 = 3*5 + b
b = - 13
Equation y = 3x - 13

Using Distance to get D
d^2 = (x1 - x2)^2 + (y2 - y1)'^2
d^2 = (x2 - 5)^2 + (y2 - 2)^2
y2 = 3*x2 - 13
10 = x2^2 - 10x2 + 25 + (3x2 - 13 - 2)^2
10 = x2^2 - 10x2 + 25 + (3x2 - 15)^2
10 = x1^2 - 10x2 + 25 + 9x^2 - 90x2 + 225
10 = 10x2^2 - 100x2 +250
0 = 10x2^2 - 100x2 + 240 Divide through by 10
0 = x2^2 - 10x^2 + 24
0 = (x2 - 6)(x2 - 4)

x2 = 6 or
x2 = 4

Using y2 = 3x - 13
for x2 = 4
then y2 = 3(4) - 13
y2 = - 1

for x2 = 6
y2 = 3*6 - 13
y2 = 5

Possible Answers for D
D = (4,-1)
D = (6,5)

I've included a graph to show a graphical solution.

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7 0

3 years ago

Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n.

Otrada [13]

I guess the "5" is supposed to represent the integral sign?

$I=\displaystyle\int_1^4\ln t\,\mathrm dt$

With $n=10$ subintervals, we split up the domain of integration as

[1, 13/10], [13/10, 8/5], [8/5, 19/10], ... , [37/10, 4]

For each rule, it will help to have a sequence that determines the end points of each subinterval. This is easily, since they form arithmetic sequences. Left endpoints are generated according to

$\ell_i=1+\dfrac{3(i-1)}{10}$

and right endpoints are given by

$r_i=1+\dfrac{3i}{10}$

where $1\le i\le10$ .

a. For the trapezoidal rule, we approximate the area under the curve over each subinterval with the area of a trapezoid with "height" equal to the length of each subinterval, $\dfrac{4-1}{10}=\dfrac3{10}$ , and "bases" equal to the values of $\ln t$ at both endpoints of each subinterval. The area of the trapezoid over the $i$ -th subinterval is

$\dfrac{\ln\ell_i+\ln r_i}2\dfrac3{10}=\dfrac3{20}\ln(ell_ir_i)$

Then the integral is approximately

$I\approx\displaystyle\sum_{i=1}^{10}\frac3{20}\ln(\ell_ir_i)\approx\boxed{2.540}$

b. For the midpoint rule, we take the rectangle over each subinterval with base length equal to the length of each subinterval and height equal to the value of $\ln t$ at the average of the subinterval's endpoints, $\dfrac{\ell_i+r_i}2$ . The area of the rectangle over the $i$ -th subinterval is then

$\ln\left(\dfrac{\ell_i+r_i}2\right)\dfrac3{10}$

so the integral is approximately

$I\approx\displaystyle\sum_{i=1}^{10}\frac3{10}\ln\left(\dfrac{\ell_i+r_i}2\right)\approx\boxed{2.548}$

c. For Simpson's rule, we find a quadratic interpolation of $\ln t$ over each subinterval given by

$P(t_i)=\ln\ell_i\dfrac{(t-m_i)(t-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+\ln m_i\dfrac{(t-\ell_i)(t-r_i)}{(m_i-\ell_i)(m_i-r_i)}+\ln r_i\dfrac{(t-\ell_i)(t-m_i)}{(r_i-\ell_i)(r_i-m_i)}$

where $m_i$ is the midpoint of the $i$ -th subinterval,

$m_i=\dfrac{\ell_i+r_i}2$

Then the integral $I$ is equal to the sum of the integrals of each interpolation over the corresponding $i$ -th subinterval.

$I\approx\displaystyle\sum_{i=1}^{10}\int_{\ell_i}^{r_i}P(t_i)\,\mathrm dt$

It's easy to show that

$\displaystyle\int_{\ell_i}^{r_i}P(t_i)\,\mathrm dt=\frac{r_i-\ell_i}6(\ln\ell_i+4\ln m_i+\ln r_i)$

so that the value of the overall integral is approximately

$I\approx\displaystyle\sum_{i=1}^{10}\frac{r_i-\ell_i}6(\ln\ell_i+4\ln m_i+\ln r_i)\approx\boxed{2.545}$

4 0

3 years ago

Erin is pouring cake batter into a pan. The pan she is using is 8 inches wide,

sergey [27]

Answer:

242.52 cubic inches

Step-by-step explanation:

Volume of the cake pan = Length × Width × Height

From the about question, we have the following dimensions for the cake pan

8 inches wide = Width

11 inches long = Length

7 cm deep = Height

We are asked to find the maximum volume in inches. Hence all the dimensions have to be in inches.

Converting Height in cm to inches

From the question,

2.54 cm = 1 inch

7cm = x inch

Cross Multiply

2.54 × x = 7 × 1

x = 7/2.54

x = 2.7559055118 inches

Volume of the cake pan =

8 × 11 × 2.7559055118

= 242.51968504 cubic inches

Approximately, the volume of the cake pan = 242.52 cubic inches

What is the maximum volume, in cubic

inches, the cake pan can hold is 242.52 cubic inches

3 0

3 years ago

What is the answer to this?

Dmitry [639]

Answer:

(c) III

Step-by-step explanation:

If you simplify the equations and the left side is identical to the right side, then there are an infinite number of solutions: the equation is true for all values of x.

Another way to simplify the equation is to subtract the right side from both sides. If that simplifies to 0 = 0, then there are an infinite number of solutions.

__

2x -6 -6x = 2 -4x . . . . eliminate parentheses

-4x -6 = -4x +2 . . . . no solutions (no value of x makes this true)

__

x +2 = 15x +10 +2x . . . . eliminate parentheses

x +2 = 17x +10 . . . . one solution (x=-1/2)

__

4 +6x = 6x +4 . . . . eliminate parentheses

6x +4 = 6x +4 . . . . infinite solutions

__

6x +24 = 2x -4 . . . . eliminate parentheses; one solution (x=-7)

8 0

2 years ago

What is 2 + -5 - -13

rosijanka [135]

Answer:

10.

The answer is 10.

5 0

3 years ago

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