Please help me! 20 points! You don't have to graph btw! I only need help with 7 and 8!

Mathematics

2 answers:

Eva8 [605]2 years ago

7 0

I can't see pictures, but I'll try to help.

This is just an example.

3x+5>2x-7

-5 -5

I subtracted 5 on both sides to cancel out it out.

3x>2x-12

-2x -2x

I subtracted the 2x to cancel it out.

x>-12

---

hope it helps

let me know if more explanation/example is needed

Mumz [18]2 years ago

6 0

Answer:

7. n>9

8. put an open dot a 9 and draw arrows to the right

9. x≤55

10. put a close dot on 55 and draw arrow to the left

move those numbers to the right or left sideeee

You might be interested in

The formula to find the period of orbit of a satellite around a planet is T^2=(4pi^2/GM)r^3 where r is the orbit's mean radius,

Natalija [7]

The answer is $r= \sqrt[3]{GMT^{2}/4 \pi^{2}}$

$T^{2} = \frac{4 \pi^{2}}{GM} r^{3}$

Move $\frac{4 \pi^{2} }{GM}$ to the other side of the equation:
$T^{2} /\frac{4 \pi^{2} }{GM} = r^{3} \\ 
T^{2} *\frac{GM}{4 \pi^{2} } = r^{3}$

Rearrange:
$r^{3} = T^{2} *\frac{GM}{4 \pi^{2} } \\ 
r^{3}= \frac{T^{2} *GM}{4 \pi^{2} } \\ 
r^{3}= \frac{GMT^{2}}{4 \pi^{2} } \\ 
r^{3} = GMT^{2}/4 \pi^{2}$

Since $x^{3}= \sqrt[3]{x}$ , then
$r^{3} = GMT^{2}/4 \pi^{2} \\ 
r= \sqrt[3]{GMT^{2}/4 \pi^{2}}$

3 0

3 years ago

Read 2 more answers

Help! Best answer gets brainliest!

liraira [26]

Answer:

Step-by-step explanation:

because the length of 12 is the same of t+1 so 11+1=12

youre welcome

6 0

2 years ago

Read 2 more answers