If by 4x(2) you mean 4x to the power 2 or 4x^2...it can be solved in the following way
f(x)=4x^2-4x-3
=4x^2+2x-6x-3
=2x(2x+2)-3(2x+2)
=(2x-3)(2x+2)
Therefore x=3/2 or x=-2/2=(-1)

8 0

3 years ago

Direct variation equation if f(x) varies directly with x and f(x)=-18 when x=3

JulsSmile [24]

F(x)=-6x
Work:
-18÷3=-6

5 0

2 years ago

Need some help on homework thank you

ddd [48]

Answer:

mabye the reflixive prop

Step-by-step explanation:

6 0

3 years ago

Evaluate the integral. (remember to use absolute values where appropriate. Use c for the constant of integration.) 5 cot5(θ) sin

ozzi

$I=5\int \frac{cos^{4}\theta }{sin\theta }\times cos\theta d\theta \\\\I=5\int \left ( 1-sin^{2}\theta \right )^{2}\times \frac{cos\theta }{sin\theta }d\theta \\put\ \sin\theta =t\\\\dt=cos\theta d\theta \\\\I=5\int\frac{t^{4}+1-2t^{2}}{t}dt\ \ \ \ \ \ \ \ \ \ \because (a-b)^2=a^2+b^2-2ab\\\\I=5\left ( \int t^{3}dt + \int \frac{1}{t} -2\int t \right )dt$

by using the integration formula

we get,

$\\I=5\left ( \frac{t^{4}}{4} +logt -t^{2}\right )\\\\I=\frac{5}{4}t^{4}+5\log t-5t^{2}+c$

now put the value of t=\sin\theta in the above equation

we get,

$\int 5\cot^5\theta \sin^4\theta d\theta=\frac{5}{4}sin^{4}\theta+5\log \sin\theta - 5sin^{2} \theta+c$

hence proved

7 0

2 years ago