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The amount of money Andrew starts with we set as X

X - 940 - 5/7(X - 940) + 240 = 2/9X

X- 5/7X -2/9X -940 +4700/7+240 = 0

X-5/7X-2/9X= 940-240-4700/7

4/63X = 200/7

4X = 1800

X = 450????

Can someone check my process because either I made a mistake during calculation or there's something wrong with the question. He can't start with 450 and spend 940.

Let k be a positive real number. The line x + y = k and the circle x^2 + y^2 = 2k + 1 are drawn. Find k so that the line is tangent to the circle.

Hello Guest!

\(x^2 + y^2 = 2k + 1\\ y=\sqrt{2k+1-x^2}\\ r^2=2k+1\\ r=\sqrt{2k+1}\)

\(x+y=k\\ y=k-x\)

Distance of the straight line \(y=k-x\)  from the origin of the coordinates.

\(r=\frac{k}{\sqrt{2}}\)

\(\dfrac{k}{\sqrt{2}}=\sqrt{2k+1}\\ \dfrac{k^2}{2}=2k+1\\k^2-4k-2=0\\ k=2\pm \sqrt{4+2}\\ k=2\pm 2.449\)

\(k=4.449\)

  !

angle R = 75.52248781º

PR = 20 * cos(R) = 5

PQ = 19.36491673

Is that a typo?

Negative 

11t - 2/3 = 4 + 6 - 3t

14t - 2/3 = 10

14t = 10+2/3

t = \({{32 \over 3} \over 14} = {32 \over 42} = {16 \over 21}\)

\({\sqrt{10} \over \sqrt{21}} * {\sqrt{35} \over \sqrt{6}} = {{√2 * √5 * √5 * √7} \over √3 * √7 * √2 * √3} = {5 \over 3}\)

Michelle made some bookmarks for fundraising day. She made 1/6 of the bookmarks on the first day and 48 bookmarks on the second day. The number of bookmarks that she made on the second day was 1/5 more than the number of bookmarks she made on the first day. How many bookmarks did Michelle make for the fundraising day?

Hello RunningMan!

\(\dfrac{x}{16}(1 +\frac{1}{5})= 48\\ \dfrac{x\cdot 6}{16\cdot 5}=48\)

\(x=640\)

640 bookmarks did Michelle make for the fundraising day.

  !

Given :

Point : ⟨ 4 , 3 , - 1 ⟩ 

Sum : ⟨ -3 ,0,2⟩

We have ,

(1) Point : ⟨ 4 , 3 , - 1 ⟩ 

       ⟨ 4 , 3 , - 1 ⟩ = 

   4i + 3j - k
 

(2) Sum : ⟨ -3 ,0,2⟩ 

          ⟨ -3 ,0,2⟩ =  

   ⟨-1 , 1 , 0 ⟩ + ⟨ -2 , -1 , 2 ⟩
 

sorry for the late response 

Let Bryan had x apples.

He put 1/3 of the apples and another 4 into basket A, i.e.

\( ({x \over 3}+4){}{}\)apples in basket A.

So, remaining apples are 

x - (x/3 + 4)=\(( {2x \over 3}- 4).{}{}\)

Now, he put1/3 of the remaining apples and another 3 into basket B, i.e. 

1/3 (2x/3 - 4) + 3 = (2x/9 - 4/3 + 3) = (2x/9 + 5/3)

apples in basket B.

This time remaining apples are 

(2x/4 - 4) - (2x/9 + 5/3) = (4x/9 - 17/3).

Now, he put 1/3 of the rest and another 3 into basket C, i.e.

1/3 (4x/9 - 17/3) + 3 = (4x/27 + 10/9) apples in basket C.

So, the remaining apples are

(4x/9 - 17/3) - (4x/27 + 10/9) = (8x/27 - 61/9).

But in the end, he had 3 apples left for himself, i.e.

(8x/27 - 61/9) = 3

⇒ (8x/27) = 3 + 61/9 =88/9

⇒ x = 88/9 * 27/8 = 33

So,

33 apples he have at first.

Find all values of z such that z^4 - 4z^2 +4 = 0.

Hello Guest!

\(Replace\ z^2\ with\ x.\)

\(x^2-4x+4=0\)

       p          q

\(x=-\frac{p}{2}\pm \sqrt{(\frac{p}{2})^2-q}\\ x=2\pm \sqrt{2^2-4}\\ x=2\)

\(Replace\ x\ with\ z ^ 2. \)

\(z^2=2\\ z=\pm \sqrt{2}\)

\(z\in \{-\sqrt{2},\sqrt{2}\}\)

\(z\in \{-1.414,1.414\}\)

  !

p = k / q^2

19==k /8^2==k / 64, solve for k

k ==64 *19 ==1,216

p =1,216 / 3^2 ==1,216 / 9 ==135 + 1/9

Note: Forget the first solution.

I would say that it is around the number 69

p == k * q^2      k==constant of proportionality

19==k * 8^2, solve for k

K==19 / 64

p ==19/64 * 3^2 ==171 / 64 ==2 + 43/64

[A + B] / 2 ==75, 

Sqrt(A * B) ==21, solve for A, B

Use substitution to get:

A==3  and  B==147

Simplify the following:
1/sqrt(2) + 1/sqrt(8)

sqrt(8) = sqrt(2^3) = 2 sqrt(2):
1/sqrt(2) + 1/(2 sqrt(2))

Rationalize the denominator. 1/sqrt(2) = 1/sqrt(2)×(sqrt(2))/(sqrt(2)) = (sqrt(2))/2:
(sqrt(2))/2 + 1/(2 sqrt(2))

Rationalize the denominator. 1/(2 sqrt(2)) = 1/(2 sqrt(2))×(sqrt(2))/(sqrt(2)) = (sqrt(2))/(2×2):
(sqrt(2))/2 + (sqrt(2))/(2×2)

2×2 = 4:
(sqrt(2))/2 + (sqrt(2))/4

Put each term in (sqrt(2))/2 + (sqrt(2))/4 over the common denominator 4: (sqrt(2))/2 + (sqrt(2))/4 = (2 sqrt(2))/4 + (sqrt(2))/4:
(2 sqrt(2))/4 + (sqrt(2))/4

(2 sqrt(2))/4 + (sqrt(2))/4 = (2 sqrt(2) + sqrt(2))/4:
(2 sqrt(2) + sqrt(2))/4

2 sqrt(2) + sqrt(2) = 3 sqrt(2):

(3 sqrt(2))/4

its 50 dollars per sofware program and 35 dolllars per video game 

I'm not sure why it popped up like this 

i agree with second answer

wait, made a mistake, a = b - 10

so 

5b - 50 + 5b + b - 5 = 495

11b - 55 = 495

11b = 550

so b = 50

a + b + c = 99

a + 5 = b - 5 ===> a = b + 10

b - 5 = 5c  ===> c = \({b - 5} \over 5\)

so b + 10 + b + \({b - 5} \over 5\) = 99

so 5b + 50 + 5b + b - 5 = 495

so 11b - 45 = 495

so 11b = 540

so b = \(540 \over 11\)

a + b + c ==99.......................(1)

[a+5] = [a-5] = [c*5].............(2), solve for a, b, c

Use substitution to get:

a==40,  b==50,   c==9

From angle chasing, we get 5x - 7 = 2x + 11.

solving that gives x = 6

so 5x - 7 = 2x + 11 = 23

so ABD = 157