Ask question

Ask question

All categories

3 years ago

12

Mark the points (10, 14), (15, 11.11), and (10, 8.22). Enter the coordinates in the input window, if you wish. Then connect the

points to form a triangle. Part A Measure the lengths of the sides of the triangle and record them in the table.

1 answer:

True [87]3 years ago

4 0

Answer:

The points form a equilateral triangle of side 5.78 units.

Step-by-step explanation:

The length of the sides of the triangle is calculated using the distance formula.

The distance between the points (x₁,y₁) and (x₂,y₂) is

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

Now,

AB = $\sqrt{(11.11-14)^{2}+(15-10)^{2}}=\sqrt{8.3521+25} = \sqrt{33.3521}=5.78$

BC = $\sqrt{(8.22-11.11)^{2}+(10-15)^{2}}=\sqrt{8.3521+25} = \sqrt{33.3521}=5.78$

CA = $\sqrt{(8.22-14)^{2}+(10-10)^{2}}=5.78$

The length of the sides of the triangle is 5.78. All the sides are equal in length. It is an equilateral triangle.

You might be interested in

Eva, the owner of eva's second time around wedding dresses, currently has five dresses to be altered, shown in the order in whic

cupoosta [38]

To answer this question, an assumption must be made, that Eva spends 8 hours a day working. If this is the case, then Eva will complete jobs w, x, and v on day one, for a total of six hours. Since the next job (y) requires 4 hours, she will spend two hours working that day, leaving 2 more hours to go on that job. The next day she will spend 2 hours finishing job y, completing it, and finish the longest job z (hours) that day. This means she had 4 jobs on day one, and 2 jobs on day 2 for and average of 3 jobs per day.
This answer assumes an 8 hour work day, and that Eva can start a job she cannot finish that day.

5 0

3 years ago

A small town's phone numbers either begin 373 or 377. How many phone numbers are available?

ser-zykov [4K]

Answer:

<u><em>There are 20,000 number available.</em></u>

Explanation:

To determine <em>how many phone numbers are available </em>you need to know how many digits the number contains.

I will assume the same number of digits for other similar questions, i.e. 7.

With 7 digits, the numbers that begin with 373 or 377 can be of the form 373XXXX or 377XXXX.

Where each X can be any digit 0 - 9. Then, there are 10 different options for each X.

Thus, there are 10×10×10×10 = 10,000 different numbers starting with 373 and other 10,000 different numbers starting with 377.

In total, there are 20,000 numbers available.

6 0

3 years ago

Help! In parallelogram LMNO below, find the length of MO

pashok25 [27]

Answer:

Step-by-step explanation:

The diagonals of a parallelogram bisect each other.

5y - 8 = 3y + 1 Add 8 to both sides

5y = 3y + 1 + 8 Subtract 3y from both sides

5y - 3y = 9 Combine

2y = 9 Divide by 2

2y/2 =9/2

y = 4. 5

5y - 8 =

5(4.5) - 8 =

22.5 - 8 =

14.5

That represents 1/2 of MO

MO = 2 * 14.5

MO = 29

5 0

2 years ago

Shureka Washburn has scores of 67, 68, 76, and 63 on her algebra tests. a. Use an inequality to find the scores she must make

mash [69]

Average=(total number)/(number of items)
given that the final exam counts as two test, let the final exam be x. The weight of the final exams on the average is 2, thus the final exam can be written as 2x because any score Shureka gets will be doubled before the averaging.
Hence our inequality will be as follows:
(67+68+76+63+2x)/6≥71
(274+2x)/6≥71
solving the above we get:
274+2x≥71×6
274+2x≥426
2x≥426-274
2x≥152
x≥76

b] The above answer is x≥76, the mean of this is that if Shureka is aiming at getting an average of 71 or above, then she should be able to get a minimum score of 76 or above. Anything less than 76 will drop her average lower than 71.

8 0

3 years ago

Samples of emissions from three suppliers are classified for conformance to air-quality specifications. The results from 100 sam

Vlad [161]

Answer

$P(A) = 0.30$

$P(B) = 0.77$

$P(A\ n\ B) = 0.22$

$P(A\ u\ B) = 0.85$

Explanation:

Given

See attachment for proper data presentation

$n = 100$ --- Sample

A = Supplier 1

B = Conforms to specification

Solving (a): P(A)

Here, we only consider data in sample 1 row.

Here:

$Yes = 22$ and $No = 8$

$n(A) = Yes + No$

$n(A) = 22 + 8$

$n(A) = 30$

P(A) is then calculated as:

$P(A) = \frac{n(A)}{Sample}$

$P(A) = \frac{30}{100}$

$P(A) = 0.30$

Solving (b): P(B)

We only consider data in the Yes column.

Here:

$(1) = 22$ $(2) = 25$ and $(3) = 30$

$n(B) = (1) + (2) + (3)$

$n(B) = 22 + 25 + 30$

$n(B) = 77$

P(B) is then calculated as:

$P(B) = \frac{n(B)}{Sample}$

$P(B) = \frac{77}{100}$

$P(B) = 0.77$

Solving (c): P(A n B)

Here, we only consider the similar cell in the yes column and sample 1 row.

i.e. [Supplier 1][Yes]

This is represented as: n(A n B)

$n(A\ n\ B) = 22$

The probability is then calculated as:

$P(A\ n\ B) = \frac{n(A\ n\ B)}{Sample}$

$P(A\ n\ B) = \frac{22}{100}$

$P(A\ n\ B) = 0.22$

Solving (d): P(A u B)

This is calculated as:

$P(A\ u\ B) = P(A) + P(B) - P(A\ n\ B)$

This gives:

$P(A\ u\ B) = \frac{30}{100} + \frac{77}{100} - \frac{22}{100}$

Take LCM

$P(A\ u\ B) = \frac{30+77-22}{100}$

$P(A\ u\ B) = \frac{85}{100}$

$P(A\ u\ B) = 0.85$

7 0

3 years ago