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At the beginning he had 66 blue + whatever number of blue ones that he gives away.

Let B be the number of Blue at the beginning

So initially he has B blue and  B+66 red

he gives away k of each so now he has

Blue:  (B-k)     and    Red:  B+66-k = (B-k)+66

We are told that he now has twice as many red so   B-k must equal 66

So    B=66+k

So he starts out with 66 blue plus the number of blue that he gives away.

So

if he gives away 5 of each then he starts with 66+5 blue and  132+5 red

if he gives away 20 of each then he starts with 66+20 blue and  132+20 red

For what values of y does the following equation have no solutions for x?
xy+6y+20x+4=1

Hello Guest!

\(xy+6y+20x+4=1 \\ x(y+20)=1-4-6y\\ x=-\dfrac{3+6y}{y+20}\)

For  y = - 20 does the equation have no solution for x.

Find the circumradius.

Hello Guest!

Cosine law:

\(cos(\alpha)=\dfrac{b^2+c^2-a^2}{2bc}=\dfrac{2\cdot 25^2-20^2}{2\cdot 25\cdot 25}=0,68\\ \alpha=acos(0,68)=47.156°\\ \color{blue}\frac{\alpha}{2}=23.578°\)

\(cos(\frac{\alpha}{2})=\dfrac{\frac{25}{2}}{r}\\ r=\dfrac{12.5}{cos(23.578°)}=13.639\)

\(The\ circumradius\ is\ \color{blue}r=13.639\).

  !

We have to find the interval of points that are less than -4

i.e.

We have to find x such that x < -4

So,

The interval notation is as follows:

x ∈ (−∞, −4)

We have to find the set of points from -5 to 2

i.e.

We have to find x such that −5 ≤ x ≤ 2

The interval notation for this is as follows 

x ∈ [−5, 2]

but

We have to exclude -1 and 2 as a union of intervals.

So,

The Interval notation becomes

x ∈ [−5, −1) ∪ (−1, 2)

Answer:

     1. (−∞, −4)

     2. [−5, −1) ∪ (−1, 2)

i got    3sqrt2/8 cis 31pi/18

check

\(\frac{\sqrt3}{2}*\frac{\sqrt6}{4}=\frac{\sqrt{2*9}}{8}=\frac{3\sqrt2}{8}\)

Angle ABD = 143 degrees.

first convert to polar form

\(2\sqrt{3}-2i\\=4(\frac{\sqrt{3}}{2}-\frac{i}{2})\\=4(\cos(-30)+i\sin(-30))\\=4(cis(-30))\)

The magnitude will be fourth-rooted, while the angle will divide by 4. 

Also, note that adding 90 to the angle will still make it a root, because 90*4=360.

Therefore the answers are:

\(\sqrt{2}(cis(-7.5)), \sqrt{2}(cis(-7.5)+90), \sqrt{2}(cis(-7.5+180)), \sqrt{2}(cis(-7.5+270))\)

The smallest n such that a_n < 1/2018 is 45.

To the nearest tenth, the area of ECODF is 3.7.

a) There are 70 multiples of 6 between 100 and 500.

c) There are 99 integers between 100 and 500 that are multiples of either 6, 8, or both.

a) There are 20 ways to divide the bracelets.

b) There are 240 ways to divide the bracelets.

Um please excuse the "imposter" this is NOT me and I profusely apologise for this and any other rude replies they have left. I am not sure how to approach this question since I am not great at those types of question:)

SCZ

LOOK https://web2.0calc.com/questions/help-thanks_64

We know that (a+b)(b-a) simplifies to b^2-a^2 so the least possible value is if b=1 and a=2:

1-4=$\boxed{-3}$

24kx = 6

x = 6/24k= 1/4k

We can use the quadratic formula:

$\frac{-17\pm\sqrt{289-88}}{4}$ are the roots. $ab=\frac{-17+\sqrt{289-88}}{4}\cdot\frac{-17-\sqrt{289-88}}{4}=\frac{289-201}{16}=\frac{11}2$

$a+b=\frac{-17+\sqrt{289-88}}{4}+\frac{-17-\sqrt{289-88}}{4}=-\frac{17}2$

$ab+a+b=\frac{11-17}{2}=\boxed{-3}$

Because the discriminant must be 0, k will have a positive and a negative that have the same absolute value(sorry for the excessive use of the word value XD), which means that the average of the two values will be $\boxed0$

Sum ==12/2 * [2*3  +  2*(12 - 1)] ==168 + 1 + [2 *9] ==187 sum of the first 100 terms.