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Find the ordered triplet (x,y,z) for the system of equations.

Hello Guest!

  \(x + 3y + 2z = 1\ |\ \times 2\) 

     -3x + y + 5z = 10

      -2x + 3y + z = 7

  \(2x+6y+4z=2\)

\(\underline{-2x + 3y + z = 7}\)

           \(9y+5z=9\ |\ \times 10\)

  \(x + 3y + 2z = 1\ |\ \times 3\\ \ \\ 3x+9y+6z=3\)

\( \underline{ -3x + y + 5z = 10}\)

       \(10y+11z=13\ |\ \times 9\)

  \(90y+50z=90\)

  \(\underline{90y+99z=117}\)

              \(49z=27\)

                 \(z=\dfrac{27}{49}\)

   \(10y+11z=13\ |\ insert\ z\)

  \(10y+\dfrac{11\cdot 27}{49}=13\\ 490y=13\cdot49-11\cdot 27=340\)

              \(y=\dfrac{34}{49}\)

  \(x + 3y + 2z = 1\\ x=1-3y-2z\ |\ insert\ y\ and\ z\\ x=1-\frac{3\cdot 34}{49}-\frac{2\cdot 27}{49}\\ 49x=49-3\cdot 34-2\cdot27\\ 49x=-107\) 

          \(x=-\dfrac{107}{49}\)

  \((x,y,z)=(-\dfrac{107}{49},\dfrac{34}{49},\dfrac{27}{49})\)

  !

How do you know those end caps aren't hemispherical? 

is it A?

120  +  130 (n-1) = 180 (n-2)

120   +  130n -130 = 180n-360

350 = 50n

n=7

At the butcher shop of a supermarket, 60 kg of beef were sold, which corresponds to 40% of the stock. How many kilograms were in stock?

Two end caps        +      lateral surface area

    pi r^2 *2             +      pi * d * h

pi (6^2) * 2            +      pi * 12 * 15

pi ( 72 + 180)

791.68 cm²

L W H  = 40

L (3W)(2H) = 40 (3)(2) = 240 gm

The box has volume 30 \(cm^3\), and can hold 40 grams of clay.

The second box has volume \(4*9*5 = 180cm^3\), so it can hold 240 grams of clay.

The line can be parameterized by (15,-5) + t(-7,8).  So for t = 2/3, the point is (31/3, 1/3).

20 * 1.25  * .6 =......                (40% off means you still pay  60%   or .6 times the original amount)

What is the approximate surface area of a cylinder that has a radius of 6 centimeters and a height of 15 centimeters? 

Approximate?  3 times 12 is 36, then that times 15 which is 360 + 18 = 540

and then just a little more because pi is a little more than 3.  Call it 550 cm².  

Calculated:  3.14 • 12 • 15 = 565 cm². 

Hmmm, my approximation wasn't bad.  Not real good, but not bad. 

_.

Vertex form of parabola

y =a (x+1)^2 + 7 

  = a ( x^2 + 2x + 1)       a*1 + 7   must equal the y -axis intercept of 18     so   a = 11

y = 11(x+1)^2 + 7

Oops, my mistake, sorry D:

Okay, they tell us that f(x) is 3, so lets just bring that into f(x)=x^2 -2x

3=x^2 -2x

We can solve by guess and check, but we can also use the quadratic formula.

Let's move 3 to the other side.

x^2 -2x -3=0

We get 3 and (-1), using the quadratic formula. If you don't know what that is, you can find how to use it on google. I hope you find this useful.

This problem specifically says BETWEEN .

Between  ALWAYS means   NOT INCLUSIVE

Pretty sure this problem has already been posted... You should find it once you search it up online, it was solved by CPhill.

I should have said inclusive, sorry about that. I guess it depends on whether the problem says inclusive or not. If this problem did, then it includes -2 and 3. Sorry about the mistake I made.

"and is divisible by ??", what?

-1   0     1    2      are integers    BETWEEN   -2 and 3

Multiply the top and the bottom   (numerator and denominator)   by   1+7i 

Wait what are you trying to find? The integers between -2 and 3 are just -2, -1, 0, 1, 2, 3. Hope this is what you are looking for.

This is just ratios.

New Age/Classical = 3/1

Classical is 3, from the text

New Age/3 = 3/1

Cross multiply

New Age = 9

Seems like fun, I'll do this later in the day! :D

It is   done like this  

\(x={4\over{(\sqrt5+1)(\root 4\of5+1)(\root 8\of5+1)(\root {16}\of5+1)}}.\\~\\ find \;(x+1)^{48}\\~\\ consider (\sqrt5+1)(\sqrt[4]{5}+1)(\root 8\of5+1)(\root {16}\of5+1)\\ let\;\; y=\sqrt[16]{5}\\ y^2=\sqrt[8]{5}\\ y^4=\sqrt[4]{5}\\ y^8=\sqrt[2]{5}\\ y^{16}=5\\ \text{so I now have}\\ (y^8+1)(y^4+1)(y^2+1)(y+1)\\ \text{when expanded out I get}\\ y^{15}+y^{14}+y^{13}+......+y^1+1\\ \text{This is the sum of a GP}\\ sum=\frac{y^{16}-1}{y-1}\\~\\ x+1=\frac{4}{the \;sum}+1\\ x+1=4\div \frac{y^{16}-1}{y-1}+1\\ x+1= \frac{4(y-1)}{y^{16}-1}+1\\ x+1= \frac{4(\sqrt[16]{5}-1)}{5-1}+1\\ x+1=\sqrt[16]{5}\\ (x+1)^{48}=(\sqrt[16]{5})^{48}\\ (x+1)^{48}=[(\sqrt[16]{5})]^{16}]^3\\ (x+1)^{48}=5^3\\ (x+1)^{48}=125\)

LaTex coding:

x={4\over{(\sqrt5+1)(\root 4\of5+1)(\root 8\of5+1)(\root {16}\of5+1)}}.\\~\\
find \;(x+1)^{48}\\~\\
consider
(\sqrt5+1)(\sqrt[4]{5}+1)(\root 8\of5+1)(\root {16}\of5+1)\\
let\;\; y=\sqrt[16]{5}\\
y^2=\sqrt[8]{5}\\
y^4=\sqrt[4]{5}\\
y^8=\sqrt[2]{5}\\
y^{16}=5\\
\text{so I now have}\\
(y^8+1)(y^4+1)(y^2+1)(y+1)\\
\text{when expanded out I get}\\
y^{15}+y^{14}+y^{13}+......+y^1+1\\
\text{This is the sum of a GP}\\
sum=\frac{y^{16}-1}{y-1}\\~\\
x+1=\frac{4}{the \;sum}+1\\
x+1=4\div \frac{y^{16}-1}{y-1}+1\\
x+1= \frac{4(y-1)}{y^{16}-1}+1\\
x+1= \frac{4(\sqrt[16]{5}-1)}{5-1}+1\\
x+1=\sqrt[16]{5}\\
(x+1)^{48}=(\sqrt[16]{5})^{48}\\
(x+1)^{48}=[(\sqrt[16]{5})]^{16}]^3\\
(x+1)^{48}=5^3\\
(x+1)^{48}=125