Answer:
<em>Set the function equal to 0</em>
Step-by-step explanation:
<u>Standard Form of the Quadratic Equation</u>
The form
is called the standard form of a quadratic equation. It can be clearly identified the terms of a second-degree polynomial equated to 0.
The equation is given in the form:
And we need to operate the expression to make it look like a standard form. The first logical step should be to set the function equal to 0 and then start to operate the resulting expression. It can be done by subtracting 8 on both sides of the equation:
Answer: Set the function equal to 0
Given:
square based pyramid = 3.5 cm base
side length = 7 cm
Base Area = (3.5cm)² = 12.25cm²
Surface Area = Base Area + 2sl = 12.25cm² + (2 * 3.5cm* 7cm)
S.A = 12.25 cm² + 49 cm²
S.A = 61.25 cm²
Y=2x-3
If u put in the values of x U will see that it equals the corresponding y value.
Answer:
+
*LN(|
|) +C
Step-by-step explanation:
we will have to do a trig sub for this
use x=a*tanθ for sqrt(x^2 +a^2) where a=2
x=2tanθ, dx= 2 sec^2 (θ) dθ
this turns 
into integral(sqrt( [2tanθ]^2 +4) * 2sec^2 (θ) )dθ
the sqrt( [2tanθ]^2 +4) will condense into 2sec^2 (θ) after converting tan^2(θ) into sec^2(θ) -1
then it simplifies into integral(4*sec^3 (θ)) dθ
you will need to do integration by parts to work out the integral of sec^3(θ) but it will turn into (1/2)sec(θ)tan(θ) + (1/2) LN(|sec(θ)+tan(θ)|) +C
then you will need to rework your functions of θ back into functions of x
tanθ will resolve back into 
(see substitutions) while secθ will resolve into 
sec(θ)= is from its ratio identity of hyp/adj where the hyp. is 
and adj is 2 (see tan(θ) ratio)
after resolving back into functions of x, substitute ratios for trig functions:
= + *LN(||) +C
Answer:
Step-by-step explanation:
Given: Function has two x-intercepts, one at (0,0) and one at (4,0)
To choose: the correct option
Solution:
x-intercepts of are the points whose x-coordinate is 0.
Consider
Put 
Put 
So, the function has two x-intercepts, one at (0,0) and one at (4,0).
