<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:
See below ↓↓
Step-by-step explanation:
<u>Finding the radius</u>
- C = 2πr
- 25 = 2 x 3.14 x r
- r = 25 / 6.28
- r = 3.98 cm ≅ 4 cm
<u />
<u>Surface Area</u>
- 4πr²
- 4 x 3.14 x 4 x 4
- 3.14 x 64
- 200.96
- <u>201 cm²</u>
<u></u>
<u>Volume</u>
- 4/3πr³ = 4πr² x r/3
- 201 x 4/3
- 67 x 4
- <u>268 cm³</u>
The answer would be A because each side is being multiplied by 3
Answer:
x=36
y=9
Step-by-step explanation:
Plug in 0 for x to find the y-intercept
0 + 4y = 36
4y = 36
y = 9
y-intercept (0, 9)
Plug in 0 for y to find the x-intercept
x + 4(0) = 36
x = 36
x-intercept (36, 0)
To find the correct function, just plug in 10 and 2 to see if the function equals zero with both numbers.
f(x)=x^2-12x+20
f(10)=100-120+20
f(10)=0 --> that's what we want, so let's check the other number
f(x)=x^2-12x+20
f(2)=4-24+20
f(2)=0 --> perfect. this function has zeros at both x=10 and x=2
The function that has zeros at both x=10 and x=2 is f(x)=x^2-12x+20.