LY = 5.2 , TY = 3.9
Δ LTN is a right triangle , ∠T = 90° , TY⊥LN
∴ TY² = LY * YN
∴ 3.9² = 5.2 * YN
∴ YN = 3.9²/5.2 = 2.925
∴ LN = LY + YN = 5.2 + 2.925 = 8.125
Δ LYT is a right triangle , ∠Y = 90°
LT² = LY² + YT² = 5.2² + 3.9² = 42.25
∴ LT = √42.25 = 6.5
Δ NYT is a right triangle , ∠Y = 90°
NT² = NY² + YT² = 2.925² + 3.9² = 23.765
∴ NT = √23.765 = 4.875
The preimeter of the rectangle LINT = 2 * (LT + TN)
= 2 * ( 6.5 + 4.875 )
= 2 * 11.375
= 22.75
Answer: THIRD OPTION.
Step-by-step explanation:
Given the following inequality:
You can follow these steps in order to solve it:
1. Apply Distributive property on the left side:
2. Add 15 to both sides of the inequality:
3. Finally, divide both sides of the inequality by 3:
The symbol is 
means "Greater than or equal to", and indicates that 7 is include in the solution.
Therefore, the solution can be expressed as:
Now you must mark this point with a closed dot on the number line and shade everything to the right.
Hi there!
Let's assume that one month is represented by the variable 'm', the amount of minutes you started with is 's', and minutes you spent is 'p'.
So, one month can be represented as 'm=s-p'.
The next month is a bit more tricky. This will incorporate 75 less minutes into the equation. 'm=s-75' can be used to represent this, as we assume that you didn't use any minutes in the first month, and that p=75 in this case.
If you found this especially helpful, I'd appreciate if you'd vote me Brainliest for your answer. I want to be able to assist more users one-on-one, as well as to move up in rank! :)
Answer:
-5/2(3x+4)<6-3x (multiply with -5/2)
-15/2x-10<6-3x (multiply with 2)
-15x-20<12-6x (change sides)
-15x+6x<12+20
-9x<12+20
-9x<32
x>-32/9
Hope this will help u :)
Answer: x = 8
(the correct answer was not provided as an option)
<u>Step-by-step explanation:</u>
![log_2(x-6)+log_2(x-4)=log_2(x)\\\\log_2[(x-6)(x-4)]=log_2(x)\\\\(x-6)(x-4)=x\\\\x^2-10x+24=x\\\\x^2-11x+24=0\\\\(x-3)(x-8)=0\\\\x=3\qquad x=8\\\\\\\text{The term after the log symbol (inside the parenthesis) must be greater than 0!}\\\\Check:\\3-6>0\ \text{FALSE --- so 3 is not a valid solution}\\\\8-6>0\ \text{TRUE}\\8-4>0\ \text{TRUE}\\8>0\ \text{TRUE --- 8 is a valid solution because it is true for all}](https://tex.z-dn.net/?f=log_2%28x-6%29%2Blog_2%28x-4%29%3Dlog_2%28x%29%5C%5C%5C%5Clog_2%5B%28x-6%29%28x-4%29%5D%3Dlog_2%28x%29%5C%5C%5C%5C%28x-6%29%28x-4%29%3Dx%5C%5C%5C%5Cx%5E2-10x%2B24%3Dx%5C%5C%5C%5Cx%5E2-11x%2B24%3D0%5C%5C%5C%5C%28x-3%29%28x-8%29%3D0%5C%5C%5C%5Cx%3D3%5Cqquad%20x%3D8%5C%5C%5C%5C%5C%5C%5Ctext%7BThe%20term%20after%20the%20log%20symbol%20%20%28inside%20the%20parenthesis%29%20must%20be%20greater%20than%200%21%7D%5C%5C%5C%5CCheck%3A%5C%5C3-6%3E0%5C%20%5Ctext%7BFALSE%20---%20so%203%20is%20not%20a%20valid%20solution%7D%5C%5C%5C%5C8-6%3E0%5C%20%5Ctext%7BTRUE%7D%5C%5C8-4%3E0%5C%20%5Ctext%7BTRUE%7D%5C%5C8%3E0%5C%20%5Ctext%7BTRUE%20---%208%20is%20a%20valid%20solution%20because%20it%20is%20true%20for%20all%7D)
