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2 years ago

The diagram shows several points and lines.

Mathematics

1 answer:

Ghella [55]2 years ago

4 0

Answer:

Option B And Option E

Step-by-step explanation:

three or more points are said to be collinear if they lie on a single straight line.

In the given figure :

Points J, K, and Q are collinear.

Through any two given points only one line can be drawn.

In the given figure:

There is only one line that can be drawn through points L and P.

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The cost of phone is reduced by 30% new cost is 116.83 what was the original price of the phone

slava [35]

Answer:

p = $116.83/0.70 = $166.90

Step-by-step explanation:

let p represent the original price. Then:

(1.00 - 0.30)p = $116.83, or

0.70p = $116.83.

Finally,

p = $116.83/0.70 = $166.90

3 0

2 years ago

What are the x- and y- coordinates of point E, which partitions the directed line segment from J to K into a ratio of 1:4? (–13,

Lana71 [14]

I added a screenshot of the complete question.

Answer:

(-7, -1)

Explanation:

The formulas given to calculated the x and y coordinates are as follows:

$x = (\frac{m}{m+n})(x_{2}-x_{1}) + x_{1}\\\\y = (\frac{m}{m+n})(y_{2}-y_{1}) + y_{1}$

Let's define the variables used:

(x₁ , y₁) are the coordinates of the first point while (x₂ , y₂) are the coordinates of the second point.

We are given that the segment is directed from J to K, therefore:

First point is J ..........> (x₁ , y₁) is (-15, -5)

Second point is K ....> (x₂ , y₂) is (25, 15)

m and n defined the portion of the partitioned segment (JE : EK). It is given that this ratio is 1:4. Therefore:

m = 1 and n = 4

Finally, let's substitute with these variables in the equations as follows:

$x = (\frac{1}{1+4})(25-(-15)) + (-15) = -7\\\\y = (\frac{1}{1+4})(15-(-5)) + (-5) = -1$

Based on the above, the coordinates of point E are (-7, -1)

Hope this helps :)

8 0

2 years ago

Read 2 more answers

Joe receives an average of 780 emails in his personal account and 760 emails in his work account each month. After changing his

Semmy [17]

Answer:

702 emails

Step-by-step explanation:

<h2>This problem bothers on depreciation of value, in this context it is Joe's email that has depreciated by 10%.</h2>

Given data

Average personal emails received monthly = $780 emails$

Average work emails received monthly $= 760 emails$

We are required to solve for the new amount of emails Joe will be receiving after changing his address, to find this value we need to solve for the depreciation of his personal mails.

After solving for the depreciation , we then need to subtract the depreciation from the initial number of mails to get the new number of mails.

let us solve for 10% depreciation.

$depreciation= \frac{10}{100} *780\\depreciation=0.1*780= 78 emails$

The new number of mails

$= initial number of mail- depreciation\\ =780-78= 702 emails$

Joe will be receiving an average of 702 emails in his personal account monthly

7 0

1 year ago

A boy had 20 cents. He bought x pencils for 3 cents each. If y equals the number of pennies left, write an equation showing the

Vesna [10]

Hey there!!

The total number of cents - 20

Cost for each pencil - 3 cents

Number of pencils bought - ' x '

y = number of pennies left

What is the domain ?

Show how y is dependent on x ..

Let's get this into an equation :

... The total cost for x pencils bought = 3x

... Number of pennies left = 20 - 3x

... y = number of pennies left

... y = 20 - 3x

Notice : If the x value changes, the y value changes too

... If x = 1 , then , y = 17; If x = 2 , then , y = 14

Hence, we could say y is dependent upon x

Domain = ?

... Remember - The total number of pennies = 20 ; hence, the total cost cannot go above 20 cents.

Hence, we will have to work with inequalities

The equation :

... 20 ≥ 3x

Divide 3 on both sides

20 / 3 ≥ x

x ≥ 6.667

Let's take this as 6

Hence , the domain will be :

D : { 1 , 2 , 3 , 4 , 5 , 6 }

Hope my answer helps!

6 0

1 year ago

Suppose again that we are counting the ways to distribute exams to TAs and it matters which students' exams go to which TAs. The

abruzzese [7]

The question is incorrect.

The correct question is:

Three TAs are grading a final exam.

There are a total of 60 exams to grade.

(c) Suppose again that we are counting the ways to distribute exams to TAs and it matters which students' exams go to which TAs. The TAs grade at different rates, so the first TA will grade 25 exams, the second TA will grade 20 exams and the third TA will grade 15 exams. How many ways are there to distribute the exams?

Answer: 60!/(25!20!15!)

Step-by-step explanation:

The number of ways of arranging n unlike objects in a line is n! that is ‘n factorial’

n! = n × (n – 1) × (n – 2) ×…× 3 × 2 × 1

The number of ways of arranging n objects where p of one type are alike, q of a second type are alike, r of a third type are alike is given as:

n!/p! q! r!

Therefore,

The answer is 60!/25!20!15!

6 0

1 year ago