<h3>Given</h3>
tan(x)²·sin(x) = tan(x)²
<h3>Find</h3>
x on the interval [0, 2π)
<h3>Solution</h3>
Subtract the right side and factor. Then make use of the zero-product rule.
... tan(x)²·sin(x) -tan(x)² = 0
... tan(x)²·(sin(x) -1) = 0
This is an indeterminate form at x = π/2 and undefined at x = 3π/2. We can resolve the indeterminate form by using an identity for tan(x)²:
... tan(x)² = sin(x)²/cos(x)² = sin(x)²/(1 -sin(x)²)
Then our equation becomes
... sin(x)²·(sin(x) -1)/((1 -sin(x))(1 +sin(x))) = 0
... -sin(x)²/(1 +sin(x)) = 0
Now, we know the only solutions are found where sin(x) = 0, at ...
... x ∈ {0, π}
Answer: a) x = 5 or -1 b) x = √3+2
c) x = -1/2 or -3/2
Step-by-step explanation:
a) (x − 2)² = 9
First step is to take the square root of both sides to eliminate the square
√ (x − 2)² = √9
x-2 = +-3
x = +3+2
x = 5 and;
x = -3+2
x = -1
x = 5 or -1
b) 3(x-2)² = 9
First we divide both sides by 3 to get;
(x-2)² = 9/3
(x-2)² = 3
Second step is to take the square root of both sides to eliminate the square
√(x-2)² = √3
x-2 = √3
x = √3+2
c) 6 = 24(x+1)²
Dividing both sides by 24, we have
6/24 = (x+1)²
1/4 = (x+1)²
Taking the square root of both sides we have
√1/4 = √(x+1)²
= +-1/2 = x+1
x = +1/2-1 = -1/2 and;
x = -1/2-1 = -3/2
x = -1/2 or -3/2
Answer:
-
Step-by-step explanation:
Two negatives create a positive.
G. The shape would be similar since dilation at the origin is just a translation and a translation doesn't affect the shape's size or shape.
h. They will still be similar because adding length to each side will just make the shape larger, and if the shape hasn't changed (which it hasn't) it will remain similar.
I. This is simple, just write an equation of the changes you could do the shape and it would be similar. You would have to explain how many you should move and from which sides (should be all sides). It is better for you to do this because there are millions of possible equations, and if you use the same one then it might be striked.
