Solving Liner Inequalities

If you solve the inequality x + 6 ≥ 9, the solution set is x ≥ 3. Graph x ≥ 3 on a number line.

soldier1979 [14.2K]

Answer:

d is the answer

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

For the circle with equation (x-2)^2+(y+3)^2=9, answer each question. a) whats coords of the center? b) what are the radius and

MrRissso [65]

Answer:

<h2>a) center (2, -3)</h2><h2>b) radius r = 3</h2><h2>c) in the attachment</h2>

Step-by-step explanation:

The standard form of an equation of a circle:

$(x-h)^2+(y-k)^2=r^2$

(h, k) - center

r - radius

We have:

$(x-2)^2+(y+3)^2=9\\\\(x-2)^2+(y-(-3))^2=3^2$

a) center (2, -3)

b) radius r = 3

c) in the attachment

4 0

3 years ago

I have no idea what the answers are

lara [203]

Hello from MrBillDoesMath!

Answer:

Top line: y = (2/3)x + 2

Bottom line: y = (2/3)x -1

Discussion:

The graph provided is hard to read but I did the best I could.

The top line appears to pass through the points (0,2) and (-3,0)

For this line

m = change y /change x = (0-2)/(-3-0) = -2/-3 = +2/3. So

y = mx + b => y = (2/3) x+ b. As the line passes through (0,2) set x = 0, y= 2 in y = (2/3)x + b =>

2 = (2/3) 0 + b => b = 2

Therefore y = (2/3)x + 2

The bottom line appears to pass through the points (0,-1) and (3,1)

For this line

m = change y /change x = (1-(-1)) /(3-0) = +2/-3. So

y = mx + b => y = (2/3) x+ b. As the line passes through (0,-1) set x = 0, y= -1 in y = (2/3)x + b =>

-1 = (2/3) 0 + b => b = -1

Therefore y = (2/3)x + -1

Thank you,

MrB

5 0

4 years ago

I need help with<br> If x = 5, find the value of LaTeX: 16-x16 − x

Nina [5.8K]

Answer:

-69

Step-by-step explanation:

16-(5)(16)-(5)

16x5=80

16-80-5

16-80=-64

-64-5=-69

3 0

3 years ago

Suppose $1000 is invested at a rate of 13% per year compounded monthly. (Round your answers to the nearest cent.)

masya89 [10]

Answer:

a. $1010.83

b.$1066.77

c. $1138.00

d.$13,269.22

Step-by-step explanation:

Given the annual rate as 13%(compounded monthly) and the principal amount as $1000.

a. #first we calculate the effective annual rate;

$i_m=(1+i/m)^m-1\\\\i_{12}=(1+0.13/12)^{12}-1=0.1380$

The compounded amount after 1 month is therefore:

$P_1=P(1+I_m)^n, n=1/12, i_m=0.1380, P=1000\\\\P_1=1000(1+0.1380)^{1/12}\\\\P_1=1010.83$

Hence, the principle after one month is $1010.83

b. The principal after 6 months:

-From a above we have the effective annual rate as 0.1380 and our time is 6 months:

$P_{6m}=P(1+i_m)^n, \ n=6m, P=1000, i_m=0.1380\\\\P_{6m}=1000(1+0.1380)^{6/12}\\\\=1066.77$

Hence, the principal after 6 months is $1066.77

c.The principal after 1 year:

-From a above we have the effective annual rate as 0.1380 and our time is 12 months:

$P_{1y}=P(1+I_m)^n, n=1/12, i_m=0.1380, P=1000\\\\P_{1y}=1000(1+0.1380)^{12}\\\\P_{1y}=1138$

Hence, the principal after 1 year is $1138.00

d. The principal after 20years:

-From a above we have the effective annual rate as 0.1380 and our time is 20yrs:

$P_{20y}=P(1+I_m)^n, n=1/12, i_m=0.1380, P=1000\\\\P_{20y}=1000(1+0.1380)^{12}\\\\P_{20y}=13269.22$

Hence, the principal after 20 years is $13,269.22

3 0

4 years ago