All Answers 358666 Answers

Do not post A O P S homework.

They will each get

\(\frac{1}{4}\;\;of\;\;\frac{1}{2}\)

what symbol is used for  'of'

I mean is it +  or  -  or  times or divide ?

Let $P(x)$ be a nonconstant polynomial, where all the coefficients are nonnegative integers. Prove that there exist infinitely many positive integers $n$  such that $P(n)$ is composite.

How about putting a little effort into presenting your question properly.

Note that \(2^a2^b2^c....=2^{a+b+c...}\)

In your case \(a+b+c\) forms an infinite geometric series, ie. \(b = ar, c = ar^2, ...etc\)

The limit of the infinite geometric series here is \(\frac{a}{1-r}\) or \(\frac{1/3}{1-1/3}\) 

You can complete the calculation.

edited, original answer was in error.   Thanks Alan.

3^1=3

3^2=9

3^3=27

3^4=81

None above are equivalent to 5 (mod7)

3^5 = 243 which is equivalent to 5 (mod7)    So    i=5

Go through the same process to find j.   

What is j?    Your turn. 

Find the value of n.

Hello Guest!

\(\frac{1}{27^{-4}}\cdot 9=81^n\\ 27^4\cdot9=81^n\\ 3^{3\cdot 4}\cdot 3^2=3^{4n}\\ 3^{3\cdot 4+2}=3^{4n}\\ 3\cdot 4+2=4n\\ 4n=14\)

\(n=3.5\)

  !

If  

F=4

R=0

and all the rest are 9

then the sum would be   9989

a/45=N+37/45               b/30=K+9/37                             (a+b)/15=what remainder         

a (mod45)=37     so     a mod15= 7     

b(mod30)=9        so     b mod 15 = 9

a+b (mod15) = (7+9)mod15    =   1 mod 15

This website is like all free help websites.

You can get polite answers or rude answer,

Correct answers or incorrect answers.

It is the individual answerer that dictates this.  Not the site.

My attitude is that it is up to the asker to assess whether the answer is of value or not.

All answers need to be checked, not just copied.

If people copy incorrect answers then it serves them right.

Let the number of fruits be x,

the number of pears be y the number of oranges be z.

y:z = 2:7

                           => y = 2/9 (3/5x)
y + z = x (1 - 2/5)     z = 7/9 (3/5x)

Apples : 2/5x

Pears: 2/15x

Oranges: 7/15x

2/5x: 2/15x: 7/15x = 6:2:7

See the solution here:  https://web2.0calc.com/questions/modular-number-theory#r6

                     Box A              Box B

Before transferred 563            187

After transferred  563 - x        187 + x

                (563 - x) - (187 + x) = 48

                             2x = 328

                                x = 164

So, in the end

        Box A                                      Box B

  563 - x = 563 - 164 = 199       187 + 164 = 351

Suppose the full price cost of a bench is $x.

then x + 2 * (1 - 25%)x + 2 = (1 - 50%)x = 385

        x + 2 * 0.75x + 2 * 0.50x = 385

                 x + 1.5x + x = 385

                               3.5x = 385

                                    x = 110 dollars

We can rewrite \(\large{{u \over v} + {v \over u} }\) as \(\large{{u^2 \over xy} + {v^2 \over xy}}\). Because these have common denominators, we can simplify furthur: \(\large{{x^2+y^2} \over {xy}}\). Now recall the identity: \(a^2+b^2=(a+b)^2-2ab\). We can then apply this here, to get: \(\large{{(u+v)^2-2uv} \over uv}\).

Using Vieta's, we know that \(u+v= -{ b\over a} = -{5 \over 3}\) and \(uv = {c \over a} = {11 \over 3} \)

Now, we have to substitute these values and simplify. 

Can you take it from here?

The answer was 1154.

By the intersecting chord theorem, we have this equation: \(AQ \times BQ = CQ \times DQ\)

Substituting what we know, we have: \(6 \times 12 = DQ \times CQ\)

Let \(CQ = x\) and \(DQ = y\). 

We can form the system: \(xy = 72\) and \(x+y = 30\)

Now, we have to solve for and take the smaller option for x. 

Can you do it from here?

Hint: You will substitute into a quadratic, and you will have to use the formula: \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

That was incorrect. Are you sure?

For the equation to have a double root, the discriminant (\(b^2-4ac\)) must equal 0. Substituting what we know, we have: \(b^2-4 \times 16 = 0 \)

Now, we have to solve for b. 

Can you take it from here?

angle b=60 degrees

tan(b)=√3

A) The largest possible degree of f(x) + ag(x) is 8.

B) The smallest possible degree of f(x) + bg(x) is 2.