Move the decimal so that there is one whole number.
0.00000482 => 4.82
We moved 6 units to the left so:
4.82 * 10^-6
Answer:
(x−3)(x−5)
Step-by-step explanation:
Let's factor x2−8x+15
x2−8x+15
The middle number is -8 and the last number is 15.
Factoring means we want something like
(x+_)(x+_)
Which numbers go in the blanks?
We need two numbers that...
Add together to get -8
Multiply together to get 15
Can you think of the two numbers?
Try -3 and -5:
-3+-5 = -8
-3*-5 = 15
Fill in the blanks in
(x+_)(x+_)
with -3 and -5 to get...
(x-3)(x-5)
We are asked in the problem to evaluate the integral of <span>(cosec^2 x-2005)÷cos^2005 x dx. The function is an example of a complex function with a degree that is greater than one and that uses special rules to integrate the function via the trigonometric functions. For example, we integrate
2005/cos^2005x dx which is equal to 2005 sec^2005 x since sec is the inverse of cos. The integral of this function when n >3 is equal to I=</span><span>∫<span>sec(n−2)</span>xdx+∫tanx<span>sec(n−3)</span>x(secxtanx)dx
Then,
</span><span>∫tanx<span>sec(<span>n−3)</span></span>x(secxtanx)dx=<span><span>tanx<span>sec(<span>n−2)</span></span>x/(</span><span>n−2)</span></span>−<span>1/(<span>n−2)I
we can then integrate the function by substituting n by 3.
On the first term csc^2 2005x / cos^2005 x we can use the trigonometric identity csc^2 x = 1 + cot^2 x to simplify the terms</span></span></span>
Liz attempted to factor 36. Since 36 = 3*12 and not 3+12, this is not the way to go. In stead of factoring 36, she should have split it into a friendly sum, like 36=30+6.
Then Step 1 should have been 27(6+30).
