MATHEMATICS COLLEGE

Can someone please help me god bless!

Answers

Answer 1

Answer:

$▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪$

let's solve :

$12x + 20y = 20$

$20y = 20 - 12x$

$y = (20 - 12x)/(20)$

$y = (20)/(20) - (12x)/(20)$

$y = 1 - (3)/(5) x$

so, the slope intercept form of given line will be :

$y = - (3)/(5) x + 1$

Answer 2

Answer:

Answer:

130

Step-by-step explanation:

sorry if wrong :-( :-( :-(

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Answers

Answer:

Option B

118

Step-by-step explanation:

$4{x}^(2) + 3x + 3$

By putting the value of x = 5

$= 4 {(5)}^(2) + 3 * 5 + 3$

= 4 × 25 + 15 + 3

= 100 + 18

= 118 (Ans)

Answer:

Step-by-step explanation:

4 × 5 to the power of 2 + 3 × 5 + 3

Plz helpp

I’ll mark you brainliest

Answers

Answer:52°

Step-by-step explanation:

x+x+14=90

Collect like terms

x+x=90-14

2x=76

Divide both sides by 2

2x/2=76/2

x=38

angle (x+14)°=(38+14)°=52° which is vertically opposite to angle PKQ.

Since Vertically opposite angles

Therefore angle PKQ=52

among a group of students 50 played cricket 50 played hockey and 40 played volleyball. 15 played both cricket and hockey 20 played both hockey and volleyball 15 played cricket and volley ball and 10 played all three. if every student played at least 1 game find the no of students and how many students played only cricket, only hockey and only volley ball

Answers

Answer:

Cricket only= 30

Volleyball only = 15

Hockey only = 25

Explanation:

Number of students that play cricket= n(C)

Number of students that play hockey= n(H)

Number of students that play volleyball = n(V)

From the question, we have that;

n(C) = 50, n(H) = 50, n(V) = 40

Number of students that play cricket and hockey= n(C∩H)

Number of students that play hockey and volleyball= n(H∩V)

Number of students that play cricket and volleyball = n(C∩V)

Number of students that play all three games= n(C∩H∩V)

From the question; we have,

n(C∩H) = 15

n(H∩V) = 20

n(C∩V) = 15

n(C∩H∩V) = 10

Therefore, number of students that play at least one game

n(CᴜHᴜV) = n(C) + n(H) + n(V) – n(C∩H) – n(H∩V) – n(C∩V) + n(C∩H∩V)

= 50 + 50 + 40 – 15 – 20 – 15 + 10

Thus, total number of students n(U)= 100.

Note;n(U)= the universal set

Let a = number of people who played cricket and volleyball only.

Let b = number of people who played cricket and hockey only.

Let c = number of people who played hockey and volleyball only.

Let d = number of people who played all three games.

This implies that,

d = n (CnHnV) = 10

n(CnV) = a + d = 15

n(CnH) = b + d = 15

n(HnV) = c + d = 20

Hence,

a = 15 – 10 = 5

b = 15 – 10 = 5

c = 20 – 10 = 10

Therefore;

For number of students that play cricket only;

n(C) – [a + b + d] = 50 – (5 + 5 + 10) = 30

For number of students that play hockey only

n(H) – [b + c + d] = 50 – ( 5 + 10 + 10) = 25

For number of students that play volleyball only

n(V) – [a + c + d] = 40 – (10 + 5 + 10) = 15

Answer:

Cricket only= 30

Volleyball only = 15

Hockey only = 25

Explanation of the answer:

Number of students that play cricket= n(C)

Number of students that play hockey= n(H)

Number of students that play volleyball = n(V)

From the question, we have that;

n(C) = 50, n(H) = 50, n(V) = 40

Number of students that play cricket and hockey= n(C∩H)

Number of students that play hockey and volleyball= n(H∩V)

Number of students that play cricket and volleyball = n(C∩V)

Number of students that play all three games= n(C∩H∩V)

From the question; we have,

n(C∩H) = 15

n(H∩V) = 20

n(C∩V) = 15

n(C∩H∩V) = 10

Therefore, number of students that play at least one game

n(CᴜHᴜV) = n(C) + n(H) + n(V) – n(C∩H) – n(H∩V) – n(C∩V) + n(C∩H∩V)

= 50 + 50 + 40 – 15 – 20 – 15 + 10

Thus, total number of students n(U)= 100.

Note;n(U)= the universal set

Let a = number of people who played cricket and volleyball only.

Let b = number of people who played cricket and hockey only.

Let c = number of people who played hockey and volleyball only.

Let d = number of people who played all three games.

This implies that,

d = n (CnHnV) = 10

n(CnV) = a + d = 15

n(CnH) = b + d = 15

n(HnV) = c + d = 20

Hence,

a = 15 – 10 = 5

b = 15 – 10 = 5

c = 20 – 10 = 10

Therefore;

For number of students that play cricket only;

n(C) – [a + b + d] = 50 – (5 + 5 + 10) = 30

For number of students that play hockey only

n(H) – [b + c + d] = 50 – ( 5 + 10 + 10) = 25

For number of students that play volleyball only

n(V) – [a + c + d] = 40 – (10 + 5 + 10) = 15

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Rachel is recording the temperature at her school every morning this week. On Monday, the temperature was 5°C. Part A: On Wednesday, Rachel calculated that the temperature was 13 degrees colder than Monday. What temperature did Rachel record on Wednesday? Part B: On Friday, Rachel calculated that the temperature was 4 degrees warmer than Wednesday. What temperature did Rachel record on Friday?

Answers

Answer:

Part A: -8

Part B: -4

Step-by-step explanation:

$5-13=-8$

$-8+4=-4$

Monticello residents were surveyed concerning their preferences for candidates Moore and Allen in anupcoming election. Of the 800 respondents, 300 support neither Moore nor Allen, 100 support both Moore and
Allen, and 250 support only Moore. How many residents support Moore or Allen?