Part 1) ![x\geq10](/tpl/images/0456/0834/d4243.png)
Part 2)
Part 3) ![p\geq 5](/tpl/images/0456/0834/0e6bd.png)
Part 4)
Part 5) ![b](/tpl/images/0456/0834/775a4.png)
Part 6) ![n](/tpl/images/0456/0834/c0919.png)
Part 7) ![n](/tpl/images/0456/0834/ba371.png)
Part 8) ![r\leq 4](/tpl/images/0456/0834/d8da7.png)
Part 9) ![x\geq 7](/tpl/images/0456/0834/197c8.png)
Part 10) ![p\leq 0](/tpl/images/0456/0834/463e8.png)
Part 11) ![x](/tpl/images/0456/0834/71305.png)
Part 12) ![a 24](/tpl/images/0456/0834/368d9.png)
Step-by-step explanation:
Part 1)
Subtract 4 both sides
![2x\geq24-4](/tpl/images/0456/0834/32bec.png)
![2x\geq20](/tpl/images/0456/0834/bf02d.png)
Divide by 2 both sides
![x\geq10](/tpl/images/0456/0834/d4243.png)
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)
Adds 3 both sides
Multiply by 3 both sides
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)
applying the distributive property left side
adds 3 both sides
Multiply by -1 both sides
![3p\geq 15](/tpl/images/0456/0834/1a3ba.png)
Divide by 3 both sides
![p\geq 5](/tpl/images/0456/0834/0e6bd.png)
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)
applying the distributive property left side
Subtract 16 both sides
Multiply by -1 both sides
Divide by 4 both sides
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-28](/tpl/images/0456/0834/3869d.png)
adds 2 both sides
![-b8+2](/tpl/images/0456/0834/cfee1.png)
![-b10](/tpl/images/0456/0834/d00e7.png)
Multiply by -1 both sides
![b](/tpl/images/0456/0834/775a4.png)
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](/tpl/images/0456/0834/ce30e.png)
applying the distributive property left side
![-12-4n-32](/tpl/images/0456/0834/b0d17.png)
adds 12 both sides
![-4n-32+12](/tpl/images/0456/0834/e3d40.png)
![-4n-20](/tpl/images/0456/0834/89ecc.png)
multiply by -1 both sides
![4n](/tpl/images/0456/0834/9548f.png)
divide by 4 both sides
![n](/tpl/images/0456/0834/c0919.png)
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}](/tpl/images/0456/0834/821af.png)
Subtract 4 both sides
![\frac{n}{3}](/tpl/images/0456/0834/96354.png)
![\frac{n}{3}](/tpl/images/0456/0834/f1e87.png)
Multiply by 3 both sides
![n](/tpl/images/0456/0834/ba371.png)
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](/tpl/images/0456/0834/1ab91.png)
applying the distributive property left side
![-3r+12\geq 0](/tpl/images/0456/0834/756c3.png)
subtract 12 both sides
divide by -1 both sides
![3r\leq 12](/tpl/images/0456/0834/07825.png)
divide by 3 both sides
![r\leq 4](/tpl/images/0456/0834/d8da7.png)
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)
Adds 7 both sides
![-7x\leq -56+7](/tpl/images/0456/0834/433ea.png)
![-7x\leq -49](/tpl/images/0456/0834/2510a.png)
Multiply by -1 both sides
![7x\geq 49](/tpl/images/0456/0834/d39ad.png)
Divide by 7 both sides
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)
applying the distributive property left side
subtract 21 both sides
Multiply by -1 both sides
![3p\leq 0](/tpl/images/0456/0834/39975.png)
![p\leq 0](/tpl/images/0456/0834/463e8.png)
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](/tpl/images/0456/0834/a9fcb.png)
Adds 4 both sides
![-11x -15+4](/tpl/images/0456/0834/dd45d.png)
![-11x -11](/tpl/images/0456/0834/d68a6.png)
Multiply by -1 both sides
![11x](/tpl/images/0456/0834/bd085.png)
Divide by 11 both sides
![x](/tpl/images/0456/0834/71305.png)
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)
Multiply by 15 both sides
![-9+a 15](/tpl/images/0456/0834/fbf5f.png)
Adds 9 both sides
![a 15+9](/tpl/images/0456/0834/23b58.png)
![a 24](/tpl/images/0456/0834/368d9.png)
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
![Can someone show work for these problems will give ! emergency](/tpl/images/0456/0834/0ac2e.jpg)
![Can someone show work for these problems will give ! emergency](/tpl/images/0456/0834/dab87.jpg)
![Can someone show work for these problems will give ! emergency](/tpl/images/0456/0834/7205a.jpg)