Answer:
-27
Step-by-step explanation:
All points in which the y-value is the 4x the x-value so some points are
(1,4)
(2,8)
(3,12)
(4,16)
(5,20)
etc.
This is due to the fact the the perimeter of a square is always 4 times any side, since all sides are equal
Answer:
Since we know that ΔPQR is a right triangle, we can also asume that:
sin R = cos P = 3/5
So the answer is (d).
* This formular can also be applied to other right triangles.
In a right triangle, sine of one acute angle will always be equal to cosine of the other acute angle.
And we can check this by actually finding cos P using the lengths of the sides, by calculating PR first:
PR = √(PQ² + RQ²) = √(12² + 16²) = 20
=> cos P = PQ/PR = 12/20 = 3/5
Answer: x = 8
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I'm going to use the notation log(2,x) to indicate "log base 2 of x". The first number is the base while the second is the expression inside the log (aka the argument of the log)
log(2,x) + log(2,(x-6)) = 4
log(2,x*(x-6)) = 4
x*(x-6) = 2^4
x*(x-6) = 16
x^2-6x = 16
x^2-6x-16 = 0
(x-8)(x+2) = 0
x-8 = 0 or x+2 = 0
x = 8 or x = -2
Recall that the domain of log(x) is x > 0. So x = -2 is not allowed. The same applies to log(2,x) as well.
Only x = 8 is a proper solution.
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You can use the change of base rule to check your work
log base 2 of x = log(2,x) = log(x)/log(2)
log(2,(x-6)) = log(x-6)/log(2)
So,
(log(x)/log(2)) + (log(x-6)/log(2)) = 4
(log(8)/log(2)) + (log(8-6)/log(2)) = 4
(log(8)/log(2)) + (log(2)/log(2)) = 4
(log(2^3)/log(2)) + (log(2)/log(2)) = 4
(3*log(2)/log(2)) + (log(2)/log(2)) = 4
3+1 = 4
4 = 4
The answer is confirmed
Answer:
<em>t=60</em>
Step-by-step explanation:
3/2t-16=4/3t-6
(subtract 4/3t from both sides)
3/2t-16-4/3t=-6
(add 16 to both sides)
3/2t-4/3t=-6+16
(simplify)
1/6t=10
(divide by 1/6 (or multiply by 6 on both sides)
t=60