Answer:
13. 
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DON'T GET HOOKED
Step-by-step explanation:
13. 
12. ![5[r + 4] = 45 \\ \\ r + 4 = 9 \\ \\ 5 = r](https://tex.z-dn.net/?f=5%5Br%20%2B%204%5D%20%3D%2045%20%5C%5C%20%5C%5C%20r%20%2B%204%20%3D%209%20%5C%5C%20%5C%5C%205%20%3D%20r)
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6. ![3[a + 5] = 24 \\ \\ a + 5 = 8 \\ \\ 3 = a](https://tex.z-dn.net/?f=3%5Ba%20%2B%205%5D%20%3D%2024%20%5C%5C%20%5C%5C%20a%20%2B%205%20%3D%208%20%5C%5C%20%5C%5C%203%20%3D%20a)
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D O N ' T G E T H O O K E D
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I am delighted to assist you anytime!
We have that
<span>3x-2y=8 -----> equation 1
2x+3y=Q----> equation 2
the solution is the point </span><span>(4,2)
in the equation 2 substitute the value of
x=4
y=2
so
</span>2x+3y=Q------> 2*4+3*2=Q-------> Q=8+6------> Q=14
<span>
the answer is
Q=14
</span>
I believe the answer is 0.008
because
0.2 × 0.2 × 0.2 is equivalent to 0.008
The square root of a a negative integer is imaginary.
It would still be a negative under a square root if you multiplied it by 2, therefor it will still be imaginary, or I’m assuming as your book calls it, undefined.
2•(sqrt-1) = 2sqrt-1
If you add a number to -1 itself, specifically 1 or greater it will become a positive number or 0 assuming you just add 1. In that case it would be defined.
-1 + 1 = 0
-1 + 2 = 1
If you add a number to the entire thing “sqrt-1” it will not be defined.
(sqrt-1) + 1 = 1+ (sqrt-1)
If you subtract a number it will still have a negative under a square root, meaning it would be undefined.
(sqrt-1) + 1 = 1 + (sqrt-1)
however if you subtract a negative number from -1 itself, you end up getting a positive number or zero. (Subtracting a negative number is adding because the negative signs cancel out).
-1 - -1 = 0
-1 - -2 = 1
If you squared it you would get -1, which is defined
sqrt-1 • sqrt-1 = -1
and if you cubed it, you would get a negative under a square root again, therefor it would be undefined.
sqrt-1 • sqrt-1 • sqrt-1 = -1 • sqrt-1 = -1(sqrt-1)
Sorry if this answer is confusing, I don’t have a scientific keyboard, I’ll get one soon.
