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but it's wrong so- 😅😅

Ah, apologies, yes, it should be 11x! Thanks :))

I think they mean the ? is also a consonant

Hi I am sorry, did you mean 5x^2 - 11x + 4 = 0? If yes, then 1/a + 1/b = 0

If not, then there's only 1 answer.

Hi Guest!

*I do not claim any credit for the answer*
 

p + q = 9

For more information:

https://web2.0calc.com/questions/algebra_18810

We can simplify the equation to \(x^2 + 2x + 10\). 

Because the leading coefficient is 1, we know \(a = 1\)

Now, notice that \(u\) must be half of the coefficient of x, meaning \(u = 1\).

Now, expand what we have, and identify what we need to add.

Can you take it from here?

nope

and what?  Is something missing?

\(ab^5=4608\) -----> 1

\(ab^4=384\) -------> 2

\(\frac{1}{2}\) Divide 1 by 2

\(b=\frac{4608}{384}\)

\(b=12\)

\(a\times (12)^4=384\)

\(a=\frac{1}{54}\)

Hi Hipie,

Might I suggest that you change your icon.  Members are given priority, by me and by many others. Your icon is so non-descript that it is easy to overlook your membership status.

Using L'hopital's Rule, Find    \(\displaystyle \lim_{x\rightarrow\infty }\frac{(x^2 + 2x + 3)}{(3x+2)^2}\)

I will work this properly in a moment but I can see straight off that the answer is goint to be 1/9

I know this because as x tends to infinity only the highest powers of x have any consequence.

So we have

\(\displaystyle \lim_{x\rightarrow\infty }\frac{(x^2 + 2x + 3)}{(3x+2)^2}=\displaystyle \lim_{x\rightarrow\infty }\frac{(x^2 +\dots)}{((3x)^2+\dots) }=\frac{1}{9}\)

ok now I will do it properly.

\(\displaystyle \lim_{x\rightarrow\infty }\frac{(x^2 + 2x + 3)}{(3x+2)^2}\\ =\displaystyle \lim_{x\rightarrow\infty }\frac{(2x + 2)}{2(3x+2)(3)}\\ =\displaystyle \lim_{x\rightarrow\infty }\frac{2x + 2}{18x+12}\\ =\displaystyle \lim_{x\rightarrow\infty }\frac{2}{18}\\ =\frac{1}{9} \)

You are very welcome. :)

Hi Slimesewer,

You will need to check my logic and my working

Thank you so much! :)

a(x^2 - 2xh + h^2) + k = 2x^2 + 6x + 11

ax^2 - 2axh + ah^2 + k = 2x^2 + 6x + 11. 

a = 2. 

2x^2 - 4xh + 2h^2 + k = 2x^2 + 6x + 11

h = 3/2

I already got the answer but I'm going to keep going for k. 

9/2 + k = 11. k = 6.5. 

k = 13/2

the form should be 2(x + 3/2)^2 + 13/2

edit: btw I already helped like 5 people about this topic.

a(x+u)(x+u) + v

a(x^2 + 2ux + u^2) + v

ax^2 + 2aux + au^2 + v = 4x^2 + 15x + 2

a, u, v are real numbers.

now we know a = 4, then substitute in. 

I think you can take it from here.

Use the formula a_n = a_1 r^n-1

a_n = (16 * 9^n-2)/(4^n-1) (common ratio) 

a_8=(16*9^6)/(4^7)

a_8 = ?

a+b+c=47.

if we simplify and combine like terms, we will get 5x^2 - 30x + 90. 

it will then = 5(x^2 - 6x)+90

if we simplify the parenthesis, it will be 5[(x-3)^2 - 9] + 90. notice how I added the - 9 because -3 * -3 = 9.

Then, I'll multiply and distrubute the 5 to the -9, which is -45

If we combine like terms, the final will be 5(x-3)^2 + 45. 5-3+45=47.

Multiply both sides by t² + 10 

That gives you           14t = 2t² + 20

Rearrange                  2t² – 14t + 20 = 0 

Factor                         (2t – 4) (t – 5) = 0 

                                       t = 2  and  t = 5 

_.

From the image provided, we see that the line forms a triangle with an area of \(6 \times 4 \div 2 = 12\).

The smaller, unshaded triangle is similar to the large triangle, with a side ratio of \(1 \over 3\). This means that area of the small triangle is \(1 \over 9\)th of the bigger triangle. 

Can you take it from here?

Would you care to look at that again guest.?

Solve the rational expression and combining terms gives you, t=5, 2

That's 5 f's so lets do it f(1) = -1/1 = -1 f(-1) = -1/-1 = 1 f(1) = -1/1 = -1 f(-1) = -1/-1 = 1 f(1) = -1/1 = -1 so f(f(f(f(f(1))))) is -1

Hi Guest,

Since s and t are directly proportional, it means if something is done to one of them, it should be done to the other too to get the symmetrical answer.

If that didn't make sense, here's an example:

Say s=3 and t=9

If s is divided by 3, t must also be divided by 3

In the end, s=1 and t=3

Now, see if you can try it for yourself!

I hope this helped

🫶

Got it thanks 🙏