3 years ago

3 out of five picks are orange . If 12 picks are orange , how many picks are there in all?

Mathematics

1 answer:

tekilochka [14]3 years ago

8 0

20 picks in total .

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Bella has a stack of 41 bills (ones, fives, and tens) in an envelope. She has 3 less tens than fives and the quantity of ones do

AleksandrR [38]

Answer:

x+(x-3)+(2*x)=41

Step-by-step explanation:

We know that the other two bills are based of of a constant value for fives. Therefore, the fives can be x. There are 3 less tens than fives, so the number of tens equals x-3. We double the number of fives to get the number of ones, so the number of ones equals 2 times x. When we add all 3 of those together, we get our total number of 41. If you want to solve, you have to solve for x, then you can plug it back into the formula. I hope this helps!

3 0

3 years ago

Use synthetic division to determine whether the number k is an upper or lower bound (as specified for the real zeros of the func

mote1985 [20]

Solution:

The given Polynomial is :

$f(x) = 5x^4 + 4x^3 - 2x^2 + 2x + 4=5( x^4 + \frac{4}{5}x^3 - \frac{2}{5}x^2 + \frac{2}{5}x + \frac{4}{5})$

By Rational Root theorem the of Zeroes of the Polynopmial are:

$\pm\frac{1}{5},\pm\frac{2}{5},\pm\frac{4}{5},\pm1,\pm2,\pm4$

But , $f(\pm\frac{1}{5},\pm\frac{2}{5},\pm\frac{4}{5},\pm1,\pm2,\pm4)\neq 0$

So, no root of this polynomial is real.

Therefore, All the four roots of Polynomial are imaginary.

So, we can't say whether the number k=2, is an upper or lower bound of the polynomial $f(x) = 5x^4 + 4x^3 - 2x^2 + 2x + 4=5( x^4 + \frac{4}{5}x^3 - \frac{2}{5}x^2 + \frac{2}{5}x + \frac{4}{5})$ .