PLEASEEEEEEEEEEEEEEEEEE HELP!!!!!!!!!

Help me please I need this ?!!!!!!

Svetradugi [14.3K]

Answer:

16x^4 y^2

Step-by-step explanation:

4 0

3 years ago

write an equation in point-slope form for the perpendicular bisector of the segment with endpoints at A(-2,2) and B(5,4)

dimaraw [331]

The equation in point-slope form for the perpendicular bisector of the segment with endpoints at A(-2,2) and B(5,4) is $y - 3 = \frac{-7x}{2}+ \frac{21}{4}$

<h3><u>Solution:</u></h3>

Given that we have to write equation in point-slope form for the perpendicular bisector of the segment with endpoints at A(-2,2) and B(5,4)

Let us first find the slope of given line AB

<em><u>The slope "m" of the line is given as:</u></em>

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Here the given points are A(-2,2) and B(5,4)

$\text {Here } x_{1}=-2 ; y_{1}=2 ; x_{2}=5 ; y_{2}=4$

$m=\frac{4-2}{5-(-2)}=\frac{2}{7}$

Thus the slope of line with given points is $\frac{2}{7}$

We know that product of slopes of given line and slope of line perpendicular to given line is always -1

$\begin{array}{l}{\text {slope of given line } \times \text { slope of perpendicular bisector }=-1} \\\\ {\frac{2}{7} \times \text { slope of perpendicular bisector }=-1} \\ \\{\text {slope of perpendicular bisector }=\frac{-7}{2}}\end{array}$

The perpendicular bisector will run through the midpoint of the given points

So let us find the midpoint of A(-2,2) and B(5,4)

<em><u>The midpoint formula for given two points is given as:</u></em>

$\text {For two points }\left(x_{1}, y_{1}\right) \text { and }\left(x_{2}, y_{2}\right), \text { midpoint } \mathrm{m}(x, y) \text { is given as }$

$m(x, y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$

Substituting the given points A(-2,2) and B(5,4)

$m(x, y)=\left(\frac{-2+5}{2}, \frac{2+4}{2}\right)=\left(\frac{3}{2}, 3\right)$

Now let us find the equation of perpendicular bisector in point slope form

The perpendicular bisector passes through points (3/2, 3) and slope -7/2

<em><u>The point slope form is given as:</u></em>

$y - y_1 = m(x - x_1)$

$\text { Substitute } \mathrm{m}=\frac{-7}{2} \text { and }\left(x_{1}, y_{1}\right)=\left(\frac{3}{2}, 3\right)$

$y - 3 = \frac{-7}{2}(x - \frac{3}{2})\\\\y - 3 = \frac{-7x}{2}+ \frac{21}{4}$

Thus the equation in point-slope form for the perpendicular bisector of the segment with endpoints at A(-2,2) and B(5,4) is found out

7 0

4 years ago

a business has recently made some upgrades to their shop that cost them $1,200. Normally, this business has a running cost of $5

DedPeter [7]

Answer:

9.23 hours to recover the money spent.

Step-by-step explanation:

50+80=130

130x 9=1170

130x9.23=1200

5 0

3 years ago

The two-way table shows the number of sport utility vehicles with certain features for sale at the car lot. a 4-column table has

Mekhanik [1.2K]

The probability that a randomly selected car with no 4-wheel drive has third-row seats is given by: Option B: 0.4

<h3>How to calculate the probability of an event?</h3>

Suppose that there are finite elementary events in the sample space of the considered experiment, and all are equally likely.

Then, suppose we want to find the probability of an event E.

Then, its probability is given as

$P(E) = \dfrac{\text{Number of favorable cases}}{\text{Number of total cases}} = \dfrac{n(E)}{n(S)}$

where favorable cases are those elementary events who belong to E, and total cases are the size of the sample space.

<h3>What is chain rule in probability?</h3>

For two events A and B, by chain rule, we have:

$P(A \cap B) = P(B)P(A|B) = P(A)P(B|A)$

where P(A|B) is probability of occurrence of A given that B already occurred.

For this case, the table given is:

Entries 4-wheel drive No 4-wheel drive Total

Third Row Seats 18 12 30

No Third Row Seats 7 28 35

Total 25 40 65

Let we take

A = event that a randomly selected car has no-4 wheel drive

B = event that a randomly selected car has third row seats

The total ways a car can be selected = 65 = n(S)

Total ways A can happen = n(A) = 40 (from the table).

Similarly, n(B) = 30

P(A) = n(A)/n(S) = 40/65

P(B) = n(B)/n(S) = 30/65

P(A∩B) = n(A∩B)/n(S) = 12/65

As by chain rule, we have:

P(A∩B) = P(A)P(B|A) = P(B)P(A|B)

We need P( A randomly selected car has three seats given that the selected car is with no 4-wheel drive)

which is symbolically P( B | A)

Thus, we use: P(A∩B) = P(A)P(B|A)

or

$12/65 = (30/65)(P(B|A))\\\dfrac{12/65}{30/65} = P(B|A)\\\\P(B|A) = \dfrac{12}{30} = \dfrac{2}{5} = 0.4$

Thus, the probability that a randomly selected car with no 4-wheel drive has third-row seats is given by: Option B: 0.4

Learn more about probability here:

brainly.com/question/1210781

3 0

2 years ago

I need help please and thank you<br> no links

Phantasy [73]

Answer:

5x-15=2x+21

5x-2x=21+15

3x=36

x=12

b= 45°

8 0

3 years ago