Y is such that 4y _ 7<= 3y and 3y <= 5y +8. What range of values of y satisfies both inequalities

Step-by-step explanation:

Given the inequality expressions

4y - 7 ≤ 3y and 3y≤5y+8

For 4y - 7 ≤ 3y

Collect like terms

4y - 3y ≤ 7

y ≤ 7

For  3y≤5y+8

Collect like terms

3y - 5y ≤ 8

-2y ≤ 8

y ≥ 8/-2

y ≥ -4

Combining both solutions

-4≤y≤7

Hence the range of values of y that satisfies both inequalities is -4≤y≤7

-4 ≤ y ≤ 7

Step-by-step explanation:

Inequalities

Solve the system of inequalities:

4y - 7 ≤ 3y       [1]

3y ≤ 5y + 8       [2]

A.

Subtracting 3y to [1]:

y - 7 ≤ 0

Adding 7:

y ≤ 7

Subtracting 5y to [2]:

-2y ≤ 8

Dividing by -2 (and flipping the sign):

y ≥ -4

The solution of the system is all the values of y greater or equal to -4 and less or equal to 7.

B.

The solution can be expressed as:

-4 ≤ y ≤ 7

-4 ≤ y ≤ 7

Step-by-step explanation:

Inequalities

Solve the system of inequalities:

4y - 7 ≤ 3y       [1]

3y ≤ 5y + 8       [2]

A.

Subtracting 3y to [1]:

y - 7 ≤ 0

Adding 7:

y ≤ 7

Subtracting 5y to [2]:

-2y ≤ 8

Dividing by -2 (and flipping the sign):

y ≥ -4

The solution of the system is all the values of y greater or equal to -4 and less or equal to 7.

B.

The solution can be expressed as:

-4 ≤ y ≤ 7

Cost of 7 gallons=$24.50

Cost of 1 gallon=24.50/7=3.5

Cost of 15 gallons=15*3.5=52.5

Cost of 15 gallons will be $52.5

The answer is in the image 

The answer is in the image 

Speed=Distance/time

Here,

distance=15m

time=1sec

speed=15/1=15m/sec

Distance=Speed*time

time=15min=15*60sec=900sec

Distance travelled in 15 min=15*900=13,500m

=13500/1000 km=13.5Km

The wood before starting =12 feet

Left wood=6 feet

Wood used till now=12-6=6 feet

Picture frame built till now= 6/(3/4)

=8 pieces

Therefore, till now 8 pieces have been made.

The wood before starting =12 feet

Left wood=6 feet

Wood used till now=12-6=6 feet

Picture frame built till now= 6/(3/4)

=8 pieces

Therefore, till now 8 pieces have been made.

30% of 420 are first year students

first year students = (30/100 )*420

=126 students

Student who are not in first year = 420-126

=294 students