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Lunna [17]
2 years ago
9

CAN SOMEONE PLEASE SHOW WORK FOR THESE PROBLEMS WILL GIVE BAINLIEST!!!!!!!!!EMERGENCY

Mathematics
1 answer:
azamat2 years ago
5 0

Answer:

Part 1) x\geq10

Part 2) m\leq -9  

Part 3) p\geq 5

Part 4) x  

Part 5) b

Part 6)   n

Part 7)  n

Part 8) r\leq 4

Part 9) x\geq 7

Part 10) p\leq 0

Part 11) x

Part 12) a > 24

Step-by-step explanation:

Part 1) 2x+4\geq24  

Subtract 4 both sides

2x\geq24-4

2x\geq20

Divide by 2 both sides

x\geq10

the solution is the interval ------> [10,∞)

The solution is the shaded area to the right of the solid line at number 10 (closed circle).

see the attached figure  

Part 2) \frac{m}{3}-3\leq -6  

Adds 3 both sides

\frac{m}{3}\leq -6+3  

\frac{m}{3}\leq -3  

Multiply by 3 both sides

m\leq -9  

the solution is the interval ------> (-∞,-9]

The solution is the shaded area to the left of the solid line at number -9 (closed circle).

see the attached figure  

Part 3) -3(p+1)\leq -18  

applying the distributive property left side

-3p-3\leq -18  

adds 3 both sides

-3p\leq -18+3  

-3p\leq -15  

Multiply by -1 both sides

3p\geq 15

Divide by 3 both sides

p\geq 5

the solution is the interval ------> [5,∞)

The solution is the shaded area to the right of the solid line at number 5 (closed circle).

see the attached figure

Part 4) -4(-4+x)>56  

applying the distributive property left side  

16-4x>56  

Subtract 16 both sides  

-4x>56-16  

-4x>40  

Multiply by -1 both sides

4x  

Divide by 4 both sides

x  

the solution is the interval ------> (-∞,-10)

The solution is the shaded area to the left of the dashed line at number -10 (open circle).

see the attached figure

Part 5) -b-2>8

adds 2 both sides

-b>8+2

-b>10

Multiply by -1 both sides

b

the solution is the interval ------> (-∞,-10)

The solution is the shaded area to the left of the dashed line at number -10 (open circle).

Part 6) -4(3+n)>-32

applying the distributive property left side  

-12-4n>-32

adds 12 both sides

-4n>-32+12

-4n>-20

multiply by -1 both sides

4n

divide by 4 both sides

n

the solution is the interval ------> (-∞,5)

The solution is the shaded area to the left of the dashed line at number 5 (open circle).

see the attached figure

Part 7) 4+\frac{n}{3}

Subtract 4 both sides

\frac{n}{3}

\frac{n}{3}

Multiply by 3 both sides

n

the solution is the interval ------> (-∞,6)

The solution is the shaded area to the left of the dashed line at number 6 (open circle).

see the attached figure  

Part 8) -3(r-4)\geq 0

applying the distributive property left side

-3r+12\geq 0

subtract 12 both sides

-3r\geq -12    

divide by -1 both sides

3r\leq 12

divide by 3 both sides

r\leq 4

the solution is the interval ------> (-∞,4]

The solution is the shaded area to the left of the solid line at number 4 (closed circle).

see the attached figure  

Part 9) -7x-7\leq -56  

Adds 7 both sides

-7x\leq -56+7

-7x\leq -49

Multiply by -1 both sides

7x\geq 49

Divide by 7 both sides

x\geq 7  

the solution is the interval ------> [7,∞)

The solution is the shaded area to the right of the solid line at number 7 (closed circle).

see the attached figure  

Part 10) -3(p-7)\geq 21  

applying the distributive property left side

-3p+21\geq 21  

subtract 21 both sides

-3p\geq 21-21  

-3p\geq 0  

Multiply by -1 both sides

3p\leq 0

p\leq 0

the solution is the interval ------> (-∞,0]

The solution is the shaded area to the left of the solid line at number 0 (closed circle).

see the attached figure  

Part 11)  -11x-4> -15

Adds 4 both sides

-11x> -15+4

-11x> -11

Multiply by -1 both sides

11x

Divide by 11 both sides

x

the solution is the interval ------> (-∞,1)

The solution is the shaded area to the left of the dashed line at number 1 (open circle).

see the attached figure

Part 12) \frac{-9+a}{15}>1  

Multiply by 15 both sides

-9+a > 15

Adds 9 both sides

a > 15+9

a > 24

the solution is the interval ------> (24,∞)

The solution is the shaded area to the right of the dashed line at number 24 (open circle).

see the attached figure

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