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17  x  21  x   26     =    ???  in^3

Omg! Thanks the  both of you!
It helped me understood' both ways!

THX, BB   !!!!

-21x^2 -11x + 40 = -x^2 - 23x + 41       rearrange    as

20x^2  - 12x  +   1    =   0         factor  as

(10x  - 1)  (2x  - 1)   =   0

Setting  each  factor to   0  and solving  for  x  gives

x   =  1/10      or      x  =  1/2

lol i alr answered them

All  of  these  are  exactly  the  same

hypotenuse  =sqrt   [  leg 1^2   +  leg2^2  ]

I'll do the first one....you  should  be  able  to  do  the rest 

6cm  =  60mm       10cm  =  100 mm 

hypotenuse  =sqrt  (60^2  + 100^2)   =  sqrt  13600  ≈   117  mm 

By De Moivres Theorme, the answer is 96.

can you show the work?

72 CDs, they sold x sweaters and 9x CDs, meaning that x+9x+80, or 10x=80. This means that they sold 8 sweaters or 72 CDs

Blue  + Red  =   18....so.....

B  +  5B  =  18

6B  =  18

B   =18 /   6    =    3  blue

Thank you so much! :)

x +  xy^2   =   250y

x  -  xy^2  =   - 240 y         add these

2x   =  10y

x  = (10 / 2) y

x  =  5y                   sub  the  into  the  first equation  for   x

5y  + 5y * y^2   =  250 y          divide through  by  5

y   + y^3  =  50y

y^3  -  49y   =  0       

y ( y^2   -  49)  =  0

y ( y + 7) (y - 7)   =   0

Solutions  for y   =

0  , -  7  ,  7

So

x  = 5(0)   = 0

x  = 5(-7)  = -35

x  =5(7)  = 35

So....the  solutions  are  (x, y)   =  (0,0 )     (-7, -35)    (7, 35)

By using the formula here are the answers: 

Top row: A - 117mm B - 103mm C - 233mm

Bottom row: A - 202mm B - 57mm C - 232mm

here is the work

Top row A: 10^2 + 6^2 = x^2  136 =x^2, or sqrt136 cm = 11.66 cm=116.6 mm=117 mm by rounding

Top row B: 9^2+5^2 = x^2 106=x^2 or sqrt106=x = 10.3cm = 103mm

Top row C: 20^2+12^2=x^2 = 544=x^2=sqrt544 = x = 23.3cm=x = 233mm

Bottom row A: 23^2=11^2+x^2 = 529=121+x^2 = x^2=408 = sqrt408=x = 20.2cm = 202mm

Bottom row B: 9^2=7^2+x^2 = 81 = 49+x^2 = x^2=32 = sqrt32=x = 5.7cm = 57mm

Bottom row C: 30^2=19^2+x^2 = 900=361+x^2 = 539=x^2 = sqrt539 = x = 23.2 cm or 232mm

Hope this helps :)

Rearrange  as

6t^2  -  47t  + 30   =   0      use   the  quadratic  formula

t     =   47  ±  sqrt   [ 47^2  -    4*6 * 30  ]              47 ±  sqrt 1489

         ____________________________   =      _______________

                     2  *  6                                                     12   

The positive  difference  between  the solutions   =      sqrt 1489  -  (-sqrt 1489)

                                                                                     _______________________  =

                                                                                                       12

2sqrt  1489

__________      =

        12                                                                                               

sqrt  1489

________

      6

https://www.jiskha.com/questions/1854755/a-basket-contains-some-white-red-and-black-ball-there-are-15-white-and-red-balls-there

answer is there

these are review problems that my tutor gave to me to do for a test coming up. He said that he found these problems and thought they would be helpful

pls help!

Find all the solutions to
\(\sqrt[3]{ \sqrt[3]{5\sqrt{2}+x}+ \sqrt[3]{5\sqrt{2}-x} }=\sqrt{2}\)

\(\small{ \begin{array}{|rcll|} \hline \sqrt[3]{ \sqrt[3]{5\sqrt{2}+x}+ \sqrt[3]{5\sqrt{2}-x} } &=& \sqrt{2} \quad & | \quad \text{cube both sides}\\ \sqrt[3]{5\sqrt{2}+x}+ \sqrt[3]{5\sqrt{2}-x} &=& \left(\sqrt{2}\right)^3 \quad & | \quad \text{cube both sides} \\ \left( \sqrt[3]{5\sqrt{2}+x}+ \sqrt[3]{5\sqrt{2}-x}\right)^3 &=& \left(\sqrt{2}\right)^9 \\ \boxed{(a+b)^3 = a^3+b^3+3ab(a+b) } \\ 5\sqrt{2}+x + 5\sqrt{2}-x +3\sqrt[3]{(5\sqrt{2}+x)(5\sqrt{2}-x)}( \underbrace{ \sqrt[3]{5\sqrt{2}+x}+ \sqrt[3]{5\sqrt{2}-x}}_{=\left(\sqrt{2}\right)^3} ) &=& \left(\sqrt{2}\right)^9 \\ 10\sqrt{2} +3\sqrt[3]{25*2-x^2}\left(\sqrt{2}\right)^3 &=& \left(\sqrt{2}\right)^9 \\ 3\sqrt[3]{50-x^2} &=& \dfrac{\left(\sqrt{2}\right)^9 -10\sqrt{2} } {\left(\sqrt{2}\right)^3} \\ 3\sqrt[3]{50-x^2} &=& \dfrac{\sqrt{2}\left( \left(\sqrt{2}\right)^8-10 \right) } {\sqrt{2} \left(\sqrt{2}\right)^2 } \\ 3\sqrt[3]{50-x^2} &=& \dfrac{\left(\sqrt{2}\right)^8-10} {\left(\sqrt{2}\right)^2 } \\ 3\sqrt[3]{50-x^2} &=& \dfrac{16-10} {4} \\ 3\sqrt[3]{50-x^2} &=& 3 \\ \sqrt[3]{50-x^2} &=& 1 \quad & | \quad \text{cube both sides} \\ 50-x^2 &=& 1 \\ x^2 &=& 49 \\ x &=& \pm 7 \\ \hline \end{array} }\)

Solutions:
\(x=-7 \\ x=7\)

Let  C  be  the  number of  chiken sandwiches

Then  15 - C  =  the  number  of tuna sandwiches

We  have

2.30C    +  2.40 (15 - C)    =   35.10          simplify

2.30C  +  36  - 2.40C   =  35.10

-10C  =  35.10  -  36

-.10C = - .90

C  = -.90  /-.10   =  9    chicken sandwiches

For some reason it wont let me upload the picture

assumed all r tuna

(15×$2.40-$35.10)÷($2.40-$2.30) = 9

The discriminant =  

9  -  4(2)(8)   =     9  -  64    =   -55

Since  the  discriminant   not  equal to 0,  this will  have  two  (non-real)  solutions