Answer:
+2x+15
Step-by-step explanation:
-3x+5x=+2x
(+2x-7) × (+8)
(+2x)(+15)
I THINK
The equivalent expression of (8x)^-2/3 * (27x)^-1/3 is 1/12x
<h3>How to evaluate the expression?</h3>
The expression is given as:
(8x)^-2/3 * (27x)^-1/3
Evaluate the exponent 8^-2/3
(8x)^-2/3 * (27x)^-1/3 = 1/4(x)^-2/3 * (27x)^-1/3
Evaluate the exponent (27x)^-1/3
(8x)^-2/3 * (27x)^-1/3 = 1/4(x)^-2/3 * 1/3(x)^-1/3
Multiply 1/4 and 1/3
(8x)^-2/3 * (27x)^-1/3 = 1/12(x)^-2/3 * (x)^-1/3
Evaluate the exponent
(8x)^-2/3 * (27x)^-1/3 = 1/12(x)^(-2/3 -1/3)
This gives
(8x)^-2/3 * (27x)^-1/3 = 1/12(x)^(-1)
So, we have
(8x)^-2/3 * (27x)^-1/3 = 1/12x
Hence, the equivalent expression of (8x)^-2/3 * (27x)^-1/3 is 1/12x
Read more about equivalent expression at
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<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}
It would be B because eaxh section either repeats itself or is close to repeating its previous part
Answer:
y-y1=m(x-x1)
Step-by-step explanation: