21. 22 23. 24. 25. PLZZ help me it’s hard for me

5. Show that the following points are collinear. a) (1, 2), (4, 5), (8,9)

Irina-Kira [14]

Label the points A,B,C

A = (1,2)
B = (4,5)
C = (8,9)

Let's find the distance from A to B, aka find the length of segment AB.

We use the distance formula.

$A = (x_1,y_1) = (1,2) \text{ and } B = (x_2, y_2) = (4,5)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(1-4)^2 + (2-5)^2}\\\\d = \sqrt{(-3)^2 + (-3)^2}\\\\d = \sqrt{9 + 9}\\\\d = \sqrt{18}\\\\d = \sqrt{9*2}\\\\d = \sqrt{9}*\sqrt{2}\\\\d = 3\sqrt{2}\\\\$

Segment AB is exactly $3\sqrt{2}$ units long.

Now let's find the distance from B to C

$B = (x_1,y_1) = (4,5) \text{ and } C = (x_2, y_2) = (8,9)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(4-8)^2 + (5-9)^2}\\\\d = \sqrt{(-4)^2 + (-4)^2}\\\\d = \sqrt{16 + 16}\\\\d = \sqrt{32}\\\\d = \sqrt{16*2}\\\\d = \sqrt{16}*\sqrt{2}\\\\d = 4\sqrt{2}\\\\$

Segment BC is exactly $4\sqrt{2}$ units long.

Adding these segments gives

$AB+BC = 3\sqrt{2}+4\sqrt{2} = 7\sqrt{2}$

----------------------

Now if A,B,C are collinear then AB+BC should get the length of AC.

AB+BC = AC

Let's calculate the distance from A to C

$A = (x_1,y_1) = (1,2) \text{ and } C = (x_2, y_2) = (8,9)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(1-8)^2 + (2-9)^2}\\\\d = \sqrt{(-7)^2 + (-7)^2}\\\\d = \sqrt{49 + 49}\\\\d = \sqrt{98}\\\\d = \sqrt{49*2}\\\\d = \sqrt{49}*\sqrt{2}\\\\d = 7\sqrt{2}\\\\$

AC is exactly $7\sqrt{2}$ units long.

Therefore, we've shown that AB+BC = AC is a true equation.

This proves that A,B,C are collinear.

For more information, check out the segment addition postulate.

7 0

3 years ago

This question has three parts. Answer the parts in order.

nalin [4]

Answer:

The area of the smallest section is $A_{1}=100yd^{2}$

The area of the largest section is $A_{2}=625yd^{2}$

The area of the remaining section is $A_{3}=250yd^{2}$

Step-by-step explanation:

Please see the picture below.

1. First we are going to name the side of the larger square as x.

As the third section shares a side with the larger square and the four sides of a square are equal, we have the following:

- Area of the first section:

$A_{1}=10yd*10yd$

$A_{1}=100yd^{2}$

- Area of the second section:

$A_{2}=x^{2}$ (Eq.1)

- Area of the third section:

$A_{3}=width*length$

$A_{3}=10yd*x$ (Eq.2)

2. The problem says that the total area of the enclosed field is 975 square yards, and looking at the picture below, we have:

$A_{1}+A_{2}+A_{3}=975yd^{2}$

Replacing values:

$100+x^{2}+10x=975$

Solving for x:

$x^{2}+10x-875=0$

$x=\frac{-10+\sqrt{100+(4*875)}}{2}$

$x=\frac{-10+\sqrt{3600}}{2}$

$x=\frac{-10+60}{2}$

$x=25$

3. Replacing the value of x in Eq.1 and Eq.2:

- From Eq.1:

$A_{2}=25^{2}$

$A_{2}=625yd^{2}$

- From Eq.2:

$A_{3}=10*25$

$A_{3}=250yd^{2}$

3 0

3 years ago

How to solve an algebraic expression by putting values of unknown

andrey2020 [161]

Answer:

Step-by-step explanation:

Algebraic expressions tend to have atleast one unknown variable in it. In order to solve the expression we need to assign a fixed value to that variable or isolate it if the expression is equal to another value. For example, in the following expression we have the variable x, if we give it a value of 5 we simply solve like a regular expression...

5x + 3

5(5) + 3

25 + 3

28

Therefore, if we swap the variable x for the value of 5 we would get a value of 28 in this algebraic expression.

7 0

3 years ago

Distances in space are measured in light-years. The distance from Earth to a star is 6.8 × 1013 kilometers. What is the distance

aalyn [17]

6.8 x 10^13 = 68,000,000,000,000
9.46 x10 ^12 = 9,460,000,000,000

6.8 x 10^13 * 1 light year / 9.46 x 10^12 = 7.188 light years.

3 0

3 years ago

Merchandise is ordered on November 10; the merchandise is shipped by the seller and the invoice is prepared, dated, and mailed b

natali 33 [55]

Answer:

November 13

Step-by-step explanation:

Following dates are given

On November 10 = Merchandise ordered

Date of an invoice prepared, dated and mailed = November 13

Date when the merchandised received by the buyer = November 18

So, the credit period begins when the invoice is prepared, dated and the mailed by the seller to the buyer as it is the evidence of that the merchandise is ordered

4 0

3 years ago