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sergey [27]
2 years ago
13

How do I do this? Help me please ​

Mathematics
1 answer:
sergiy2304 [10]2 years ago
4 0

Answer:

4.6 and 3.5 and 3.6.

Step-by-step explanation:

Because they have to be in a range between 3.7 and 4.3.

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Find the equation of the line passing through the points (0; 2) and (1; 5)​
Charra [1.4K]

Answer:

y=3x+2

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Amanda exercised for 10 minutes every day in the first week, 20 minutes in the second week, 30 minutes in the third week, and 40
Misha Larkins [42]
                  Amanda           Billy
1st week       10                   5
2nd week      20                  10 
3rd week       30                  20
4th week       40                  40
<span>
A) Amanda's method is linear because the number of minutes increased by an equal number every week.</span>
common difference is 10.
1st week      0 + 10 = 10
2nd week   10 + 10 = 20
3rd week    20 + 10 = 30
4th week    30 + 10 = 40

Billy's method is exponential:
5(2)^x

1st week   5(2⁰) = 5(1) = 5
2nd week  5(2¹) = 5(2) = 10
3rd week   5(2²) = 5(4) = 20
4th week   5(2³) = 5(8) = 40
8 0
3 years ago
Read 2 more answers
1/10 is one part of the whole(number)divided into.......equals part.
Tom [10]
10 equal parts. that is the answer
5 0
3 years ago
Alicia used 4 gallons of gasoline to travel 90 miles. At the same rate, how far can she travel on a full tank that holds 18 gall
MrRa [10]
90 miles/4 gallons = 22.5 miles/1 gallon.

22.5 miles*18 gallons= 405 miles

the answer is 405 miles
6 0
3 years ago
There are three clubs at Redwood High School: Debate, Student council, and Key club. There are 1000 students in the school. And
hram777 [196]

Answer:

The number of students in the key club is 490

Step-by-step explanation:

The given parameters are;

The number of student in the school = 1000

The total number of student in the debate club = 310

The total number of student in the student council = 650

The total number of student who are in debate and student council = 170

The total number of student who are in both debate and the key club = 150

The total number of student who are in both student council and the key club = 180

The number of students who are in all three clubs = 50

Therefore, we have;

Let A represent the number of students in the debate club

Let B represent the number of students in the student council

Let C represent the number of students in the key club

We have;

n(A∪B∪C) = n(A) + n(B) + n(C) -  n(A∩B) - n(B∩C) - n(C∩A) + n(A∩B∩C)

Where;

n(A∪B∪C) = 1000

n(A) = 310

n(B) = 650

n(A∩B) = 170

n(B∩C) = 180

n(C∩A) = 150

n(A∩B∩C) = 50

Therefore;

n(C) = n(A∪B∪C) - (n(A) + n(B) -  n(A∩B) - n(B∩C) - n(C∩A) + n(A∩B∩C))

Substituting the values gives;

n(C) = 1000 - (310 + 650 -  170 - 180 - 150 + 50) = 490

The number of students in the key club, n(C) = 490.

5 0
3 years ago
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