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DanielleElmas [232]
3 years ago
10

A line with a slope of 4 passes through the point (2, 1). What is the equation of this line?

Mathematics
1 answer:
olasank [31]3 years ago
7 0

Answer:

point slope form: y - 1 = 4 (x -2)

slope intercept form: y = 4x -7

Step-by-step explanation:

If you want it in point slope form it would be: y - 1 = 4 (x -2)

but if you want it in slope intercept form, you solve this equation

y -1 = 4 (x-2)

y-1 = 4x -8

 +1          +1

y = 4x -7 in slope intercept form

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Prove that: (12^13–12^12+12^11)(11^9–11^8+11^7) is divisible by 3, 7, 19, and 37. The answer should be like: x^b*3*7*19*37. Also
n200080 [17]

Answer:

Step-by-step explanation:

Given (12^13–12^12+12^11)(11^9–11^8+11^7),

(12^13–12^12+12^11)(11^9–11^8+11^7) =

[(12^12)12 – 12^12 + 12^11][(11^8)11 – 11^8 + 11^7)

[(12^12)(12 – 1) + 12^11][(11^8)(11 – 1) + 11^7] =

(12^12(11) + 12^11)(11^8(10) + 11^7) =

(12^11(12x11) + 12^11)(11^7(11x10) + 11^7) =

[(12^11)(12x11 + 1)][(11^7)(11x10 + 1)] =

[(12^11)x(11^7)](12x11 + 1)(11x10 + 1) =

[(12^11)x(11^7)](133 x 111) =

[(12^11)x(11^7)](133 x 111) =

[(12^11)x(11^7)](14763) =

[(12^11)x(11^7)](3x7x19x37)

From here, it is clear that the given number is divisible by 3, 7, 19 and 37.

6 0
3 years ago
Crayons come in packs of 25 and markers come in packs of 10. Marcus wants to have the same amount of markers and crayons. What i
Trava [24]

Answer:

5 packs of markers and 2 packs of crayons

Step-by-step explanation:

they will both have 50

4 0
3 years ago
2.38 Baggage fees: An airline charges the following baggage fees: $25 for the first bag and $30 for the second. Suppose 54% of p
Anna71 [15]

The average baggage-related revenue per passenger is $16.30 per passenger.

<h3>Expected value</h3>

Expected value formula: x×p(x)

First step

No passenger=0×.54

No passenger=0

Second step

One checked luggage for first bag=.30×$25

One  checked luggage for first bag=$7.50

Third step

Two piece  for the first and second bag=.16×($25+$30)

Two piece  for the first and second bag=.16×$55

Two piece  for the first and second bag=$8.80

Last step

Expected value=$7.50+$8.80

Expected value=$16.30

Therefore the average baggage-related revenue per passenger is $16.30 per passenger.

Learn more about expected value here:brainly.com/question/24305645

#SPJ1

7 0
2 years ago
Solve y = x + 8 for x.<br> x= y + 8<br> x=y - 8<br> x= -y + 8<br> x = -y - 8
Anna007 [38]
X = y - 8

Step by step
Y = x + 8
x + 8 = y
-8 -8
x = y - 8

3 0
3 years ago
All the fourth-graders in a certain elementary school took a standardized test. A total of 85% of the students were found to be
Aneli [31]

Answer:

There is a 2% probability that the student is proficient in neither reading nor mathematics.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a student is proficient in reading

B is the probability that a student is proficient in mathematics.

C is the probability that a student is proficient in neither reading nor mathematics.

We have that:

A = a + (A \cap B)

In which a is the probability that a student is proficient in reading but not mathematics and A \cap B is the probability that a student is proficient in both reading and mathematics.

By the same logic, we have that:

B = b + (A \cap B)

Either a student in proficient in at least one of reading or mathematics, or a student is proficient in neither of those. The sum of the probabilities of these events is decimal 1. So

(A \cup B) + C = 1

In which

(A \cup B) = a + b + (A \cap B)

65% were found to be proficient in both reading and mathematics.

This means that A \cap B = 0.65

78% were found to be proficient in mathematics

This means that B = 0.78

B = b + (A \cap B)

0.78 = b + 0.65

b = 0.13

85% of the students were found to be proficient in reading

This means that A = 0.85

A = a + (A \cap B)

0.85 = a + 0.65

a = 0.20

Proficient in at least one:

(A \cup B) = a + b + (A \cap B) = 0.20 + 0.13 + 0.65 = 0.98

What is the probability that the student is proficient in neither reading nor mathematics?

(A \cup B) + C = 1

C = 1 - (A \cup B) = 1 - 0.98 = 0.02

There is a 2% probability that the student is proficient in neither reading nor mathematics.

6 0
3 years ago
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