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tomatoes in ingredients = tomatoes in final sauce

Amount of 70% = x

amount of 40% =  (6-x)

70 % (x)    +   40% (5 L - x)     =   60% ( 5 L)

.7 x           + .4 (5-x)                = .6 (5)                       SOlve for 'x' the amount of 70%           then 40% = 5-x

L = 25

M = 10 +25 = 35

A = 3 * 35 = 105

I think you should try it for the rst of the Q's from here......

The point    -4 ,   3 sqrt5     forms a right triangle with the x axis     in the 2nd quadrant  ( so cos will be NEGATIVE)

   find the hypotenuse of this right triangle

      (-4)^2 +(3 sqrt5)^2   = hyp ^2

      16      +  45              = hyp^2

              sqrt 61   = hyp

Now   cos = adj/hyp     =    - 4 / sqrt 61       =     -4 sqrt61 / 61

what does 74x7x087x37x6x5x3x76x545x674x334x666x666x443

aka x is not times I am in 6 grade

74 x 7 x 087 x 37 x 6 x 5 x 3 x 76 x 545 x 674 x 334 x 666 x 666 x 443     

I added spaces so it'd be easier to read.  These could be measurements   

of a 14-dimensional object.  You know, like width x height x depth are the    

measurements of a 3-dimendional object.  What d'ya think?    

_.

A rock is thrown upward with a velocity of 22 meters per second from the top of a 23 meter high cliff, and it misses the cliff on the way back down. When will the rock be 10 meters from ground level?

Awww, thank you!

:) :) :) :) :) :) :) :) :) :) :) :) :) 

x/24 = 3/8

     x = 3*24/8

     x = 9

Two decks contain 24 face cards     out of 104 cards total

   you rec'd 2  non-face cards ...there are 102 cards left   and 24 are face cards    24/102 = 12/51 = 4/17

sorry, both of those are wrong

The probability is (24*1 + 18*2)/C(11,4) = 2/11.

40 hours at $ 10/hr  =  $ 400

8 hours at  ($ 10 x 1.25)  = $100      add together for answer.....

If   'x' does not equal multiplication, your question is nonsensical.     

This is the answer they gave, which I got 2nd try:
 

The prime factorization of 1200 is  2^4 * 3 * 5^2. The only way to get a factor of 5 is a roll of 5 itself, so two of the rolls must have been a 5. That means the product of the other three rolls were 

The only ways to get a factor of 3 is through the rolls of 3 or 6. If one of the three remaining rolls is a 3, then the product of the other two rolls is 16. The only way this can happen is if the other two rolls were 4 and 4. If one of the three remaining rolls is a 6, then the product of the other two rolls is 8. The only way this can happen is if the other two rolls were 2 and 4.

So, the rolls must have been either 5, 5, 3, 4, 4 in some order, or 5, 5, 6, 2, 4 in some order. For 5, 5, 3, 4, 4, there are 5!/(2!*2!) =30
possible sequences. For 5, 5, 6, 2, 4, there are 5!/2! =60
possible sequences. This gives us a total of 90 possible sequences.

Let  x  =  number of basic models.

Let  y  =  number of upgraded models.

Sales:      60x + 200 y  =  1800                                          60x + 200y  =  1800

Purchase:  10x + 50y  =  400               --->   x -4   --->     -40x  - 200y  =  -1600

Adding down the columns:                                                 20x             =  200

                                                                                                          x  =  10     <---  number of basic models

Substituting:  10x + 50y  =  400     --->     10(10) + 50y  =  400     --->     100 + 50y  =  400

                                                               --->     50y  =  300     --->     y  =  6     <---   number of upgraded models

maybe it seems that way because YOU are here!

My teacher ask if 4-7 was 3 and i said yes and it wrong nooo       

I assume your teacher explained WHY that answer was incorrect?    

4 – 7 = –3      You had the "three" part right, but it's a negative three.   

Think about if you had 4 dollars, and bought something that cost 7.

You wouldn't have 3 dollars left in your pocket, you would still owe 3.  

In bookkeeping, they call that a negative balance.

_.

Karl is a thrifty boy. He always spends money wisely. He has saved $459 so far.

Alexa has  \(\frac{5}{9}\)  of Karl's money. How much must Karl give Alexa so that both will have the same amount?

Hello Guest!

\(k=495\\ a=\frac{5}{9}\cdot495\\ \frac{x}{2}=k-a=495-\frac{5}{9}\cdot495=\frac{4}{9}\cdot 495=220\\ \)

\(x=110\)

$110 must Karl give Alexa so that both will have the same amount.

  !

Push the two walls shown in CalculatorUser's picture outward until they lie flat on the ground. Then the "x-direction" distance between the two points is 7 + 12 + 1 or 20. The "y-direction" distance is 10 - 2 or 8. So the straight line between the points is given by Pythagoras as sqrt(20² + 8²) or 4sqrt(29)

Let  the  intersection point of the three chords  = A

Let  the other end of the chord of length 10 =  B

Let the other end  of the chord of length 12 = C

Connect BC

Let the intersection of BC and chord "x"  = D

Let the other end of chord  "x"  = E

Using the Law of Cosines

BC = sqrt  [10^2 + 12^2  - 2 (10*12)cos (60°)]  =  sqrt ( 244 - 120)   = sqrt (124) = 2sqrt (31)

And because angle BAC is bisected  then  in triangle ABC

BA / CA  =  BD / CD

10/12  = BD /CD

5/6 = BD / CD

So   BD =  (5/11)* BC =  (5/11)*2sqrt (31)  = (10/11) sqrt (31)

And CD = (6/11) * BC = (6/11) * 2sqrt (31)  = (12/11)sqrt (31)

And  using the Law of Sines

sin ABC / 12  =  sin 60 / [ 2sqrt (31)]

sin ABC  =  12 sqrt (3) / [ 4sqrt (31) ]

sin ABC = 3sqrt (3) / sqrt (31)

And   using the Law of Sines once more

AD / sin ABC = BD / sin BAD

AD / [ 3sqrt (3) /sqrt (31) ]  =  (10/11)sqrt (31) / sin 30

AD sqrt (31) / [ 3sqrt (3) [ =  2 *(10/11) sqrt (31)

AD  =  60 sqrt (3)  / 11  =  (60/11)sqrt (3)

And  using the intersecting chord theorem

AD * DE =  BC * CD

(60/11)sqrt (3)  * DE  =  (10/11)sqrt (31) * (12/11)sqrt (31)

(60/11) sqrt (3) * DE  =  3720 / 121

(660/121)sqrt (3) * DE =  3720/121

DE  =  (3720/121) (121/660) / sqrt (3)

DE =  (62/11) / sqrt (3)

DE =  (62/33)sqrt (3)

So   "x"  =   AD  + DE =    sqrt(3)  ( 60/11 + 62/33)   = sqrt (3) ( 180 + 62) / 33  = 

sqrt ( 3)  (242 / 33 )  =   sqrt (3) (22/3)   =  

22 / sqrt (3)

Thank you, Alan !

What does that even mean?

Solve for x.

\(\begin{array}{|rcll|} \hline \mathbf{\text{cos-rule:}} \\ \hline \mathbf{z^2} &=& \mathbf{10^2+12^2-2*10*12*\cos(60^\circ)} \quad | \quad \cos(60^\circ) = \dfrac{1}{2} \\ z^2 &=& 100+144-240*\dfrac{1}{2} \\\\ z^2 &=& 124 \\\\ z^2 &=& 4*31 \\\\ \mathbf{z} &=& \mathbf{2\sqrt{31}} \qquad (1) \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{\text{sin-rule:}} \\ \hline \mathbf{\dfrac{\sin(\alpha)}{10}} &=& \mathbf{\dfrac{\sin(60^\circ)}{z}} \\\\ \sin(\alpha) &=& \dfrac{10\sin(60^\circ)}{z} \quad | \quad \sin(60^\circ) = \dfrac{\sqrt{3}}{2} \\\\ \sin(\alpha) &=& \dfrac{10\sqrt{3}}{2z} \\\\ \sin(\alpha) &=& \dfrac{5\sqrt{3}}{z} \quad | \quad z=2\sqrt{31} \\\\ \sin(\alpha) &=& \dfrac{5\sqrt{3}}{2\sqrt{31}} \\\\ \sin(\alpha) &=& \dfrac{5}{62}\sqrt{93} \\\\ \sin(\alpha) &=& 0.77771377105 \\\\ \alpha &=& \arcsin\left(0.77771377105\right) \\\\ \mathbf{\alpha} &=& \mathbf{51.0517244354^\circ} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{\text{sin-rule:}} \\ \hline \mathbf{\dfrac{\sin(\alpha)}{10}} &=& \mathbf{ \dfrac{\sin(\beta)}{12}} \\\\ \sin(\beta) &=& \dfrac{12}{10}\sin(\alpha) \\\\ \sin(\beta) &=& \dfrac{6}{5}\sin(\alpha) \\\\ \sin(\beta) &=& \dfrac{6}{5}*0.77771377105 \\\\ \sin(\beta) &=& 0.93325652526 \\\\ \beta &=& \arcsin\left(0.93325652526\right) \\\\ \mathbf{\beta} &=& \mathbf{68.9482755646^\circ} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{\text{sin-rule:}} \\ \hline \mathbf{\dfrac{\sin(30^\circ+\alpha)}{x}} &=& \mathbf{ \dfrac{\sin(\beta)}{12}} \\\\ x &=& \dfrac{12\sin(30^\circ+\alpha)^\circ)}{\sin(\beta)} \\\\ x &=& \dfrac{12\sin(30^\circ+51.0517244354^\circ)}{0.93325652526} \\\\ x &=& \dfrac{12\sin(81.0517244354)}{0.93325652526} \\\\ x &=& \dfrac{12*0.98782916115}{0.93325652526} \\\\ x &=& \dfrac{11.8539499338}{0.93325652526} \\\\ \mathbf{x} &=& \mathbf{12.7017059222} \\ \hline \end{array} \)

This problem is neatly solved by Presh Talwalker on YouTube. Search for "Mind Your Decisions. The chord progression problem."