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What is 9x + 81 mod 19 congruent to?

81/19 = 4+5/19  so

\(9x + 81 \mod 19\; \equiv \;9x + 5 \mod 19\)

You are right guest, I made the maths much more messy than necessary.   :)

I am not going to do the whole thing for you, but setting it out in a table like this should help you.

If the roots of the equation are a and b, then

a + b = 11/5, and ab = 4/5.

So,

\(\displaystyle \frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}=\frac{11}{5}.\frac{5}{4}=\frac{11}{4}.\)

I think it looks ok except L can be positive or negative.  So you need the +/- sign

Are you sure you have copied this down properly Juriemagic?

Hi juriemagic,

it is solved correctly.
You can still put the W under the root.

\(L=\frac{\sqrt{Z^2-R^2}}{W}\)

  !

HCF =1

The rest looks good :)

A mixture of purple paint contains 6 teaspoons of red paint and 15 teaspoons of blue paint. To make the same shade of purple paint using 35 teaspoons of blue paint, how much red paint would you need?

Here is an easy way to do these without algebra

\(15*\frac{35}{15}=35\)   so I will have to multiply the red by  35/15 too.

Ask what you want.

One question per post and only 1 or 2 posts running concurrently.

If you show your own logic as a part of the question it will make a good impression and you will be more likely to get a proper answer.

The worst that happens is that someone rebukes you or no one answers you. You can handle that, can't you?

\(If \qquad 5x^2 - 11x + 4 = 0\\ \text{and a and b are the solutions then}\\ a=\frac{11+\sqrt{41}}{10}\qquad and \qquad b=\frac{11-\sqrt{41}}{10}\\~\\ \frac{1}{a}+\frac{1}{b}=\frac{10}{11+\sqrt{41}}+\frac{10}{11-\sqrt{41}}\\~\\ \frac{1}{a}+\frac{1}{b}=\frac{10(11-\sqrt{41})+10(11+\sqrt{41})}{(11+\sqrt{41})(11-\sqrt{41})}\\~\\ \frac{1}{a}+\frac{1}{b}=\frac{(110-10\sqrt{41})+(110+10\sqrt{41})}{(11+\sqrt{41})(11-\sqrt{41})}\\~\\ \frac{1}{a}+\frac{1}{b}=\frac{(110-10\sqrt{41})+(110+10\sqrt{41})}{(11+\sqrt{41})(11-\sqrt{41})}\\~\\ \)

I think you can finish it from there.

LaTex:

If \qquad 5x^2 - 11x + 4 = 0\\ \text{and a and b are the solutions then}\\

a=\frac{11+\sqrt{41}}{10}\qquad or \qquad b=\frac{11-\sqrt{41}}{10}\\~\\

\frac{1}{a}+\frac{1}{b}=\frac{10}{11+\sqrt{41}}+\frac{10}{11-\sqrt{41}}\\~\\

\frac{1}{a}+\frac{1}{b}=\frac{10(11-\sqrt{41})+10(11+\sqrt{41})}{(11+\sqrt{41})(11-\sqrt{41})}\\~\\

\frac{1}{a}+\frac{1}{b}=\frac{(110-10\sqrt{41})+(110+10\sqrt{41})}{(11+\sqrt{41})(11-\sqrt{41})}\\~\\

\frac{1}{a}+\frac{1}{b}=\frac{(110-10\sqrt{41})+(110+10\sqrt{41})}{(11+\sqrt{41})(11-\sqrt{41})}\\~\\

Dragan is correct

\(\sqrt{16}=+4\qquad \text{By convention}\\~\\ \\But\\ if\;\;y^2=16\\then\\ y=\pm\sqrt{16}=\pm4 \)

So if the sqrt root sign or the power of 1/2 is already in the question then the answer is positive (by convention).

But

If you add the square root in, as a part of the solution THEN the answer is positive OR negative.

That's incorrect srry

1

1+3=4

1+3+5=9

1+3+5+7=16

1+3+5+7+9=25

1+3+5+7+9+1=26

1+3+5+7+9+1+3=29

1+3+5+7+9+1+3+5=34

1+3+5+7+9+1+3+5+7=41

1+3+5+7+9+1+3+5+7+9=50

First number of row,    1,3,7,13,   21

So it is the 10th row

Because they are not at all helpful.

You give a number, which may or may not be correct. 

You give no explanation, your answer has no value whatsoever.

So, like other garbage, it gets deleted.

A ratio is a way to compare two quantities by using division as in miles per hour where we compare miles and hours. A proportion on the other hand is an equation that says that two ratios are equivalent. ... If one number in a proportion is unknown you can find that number by solving the proportion.

1
3, 5
7, 9, 11
13, 15, 17, 19.
21, 23. 25, 27, 29
31, 33, 35, 37, 39, 41
43, 45, 47, 49, 51, 53, 55
57, 59, 61, 63, 65, 67. 69, 71
73, 75, 77, 79, 81, 83, 85, 87, 89,
91, 93, 95, 97, 99, 101, 103, 105, 107, 109

so 10

I know. 1, 3, 5, 7, and 9 are the repeating units digits.

it says sum of units digit

IDK but the answer is 10

1+3+5+7+9 is 25 which makes it a lot easier.

why have you deleted all the answers?!? 

So is the answer 74 or 148?