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2x + 1y         =       3.15      →       x  =  3.15 - y

        1y + 1z =       3.50     →        z  =  3.50 - y

1x + 2y + 3z =       8.15

1(3.15 -y)  + 2y + 3(3.50 - y)  = 8.15

3.15 - y + 2y + 10.50 - 3y =  8.15

-2y  =  8.15 - 3.15 - 10.50

-2y = - 5.50

y = 5.50 / 2  =   2.75   =   275 cents

https://web2.0calc.com/questions/rectangle_45

x^2+0x-1. I got this by multiplying (x+1)(x-1) to get x^2-1

Let P be the number of pages

(1/5)P  = 1st day

20 = second day

(1/6)( P - (1/5)P -20)  = (1/6) ((4/5)P - 20)   =  (2/15)P - 10/3  = third day

(1/2)P  = fourth day

(1/5)P + 20 + (2/15)P -10/3 = (1/2)P       multiply through by 60

12P + 1200 + 8P - 200 = 30P

1200 - 200 =  30P -20P

1000 = 10P                         

P  = 1000 / 10 =   100  pages

A45

12/sqrt 2                   D    12

B 90           12/sqrt 2                     C 45

By AAS, triangle ABD  congruent to triangle CBD

[ ABC]  = (1/2)(12/sqrt 2)^2  = (1/2)(144/2)  = 144/4 =  36

[ABD]  = (1/2)[ABC]  = 18

xy  = x + y   →  xy - x  = y →  x ( y - 1)  = y →  x = y / (y  - 1)

xz =  x +  z

yz =  y + z   → yz - z = y → z (y - 1) = y  → z = y/(y - 1)

x = z

So

xz = x + z

z*z  = z + z

z^2  = 2z

z^2 - 2z = 0 

z ( z  - 2)  = 0 

z = 0    reject

z = 2 

Cross multiplying, we get

\(12a+24=ab+b\)

Moving ab to the other side and factoring, we have

_{\(a\left(12-b\right)=b-24\)}

Dividng both sides 12-b, we get

\(a=\frac{b-24}{12-b}\)

So, everything works, and we have \(a=\frac{b-24}{12-b};\quad \:b\ne \:12\)

So technically, there are infinite solutions. 

Thanks! :)

Combining all like terms and moving all terms to one side we get

\(6x^2-21x-31=0\)

Now, we simply use the quadratic equation, we get

\(x_{1,\:2}=\frac{-\left(-21\right)\pm \sqrt{\left(-21\right)^2-4\cdot \:6\left(-31\right)}}{2\cdot \:6}\)

\(x_1=\frac{-\left(-21\right)+\sqrt{1185}}{2\cdot \:6},\:x_2=\frac{-\left(-21\right)-\sqrt{1185}}{2\cdot \:6}\)

\(x=\frac{21+\sqrt{1185}}{12},\:x=\frac{21-\sqrt{1185}}{12}\)

So our final answers are \(x=\frac{21+\sqrt{1185}}{12},\:x=\frac{21-\sqrt{1185}}{12}\)

Thanks! :)

First, let's simplfy our equation. We have

\(2x^2 - 3x - 35 = -43 + jx \\ 2x^2 - 3x - jx + 8 = 0 \\ 2x^2 - (3 + j)x + 8 = 0\)

If there is only one solution, then the descriminant of the quadratic must be 0. So, we can write the equation

\((3 + j)^2 - 4(2)(8) = 0 \\ (j + 3)^2 - 64 = 0 \\ [ (j + 3) - 8 ] [ (j + 3) + 8 ] = 0 \\ [ j - 5 ] [ j + 11] = 0\)

Thus, we have that 

\(j = 5 \\ j = -11\)

So our final answer is 5 and -11. 

Thanks! :)

Let's first combine all like terms. We have

\(x^2+kx-9x+16\\ x^2+(k-9)x+16\)

Since we need this to be a difference of squares, k-9 must be 2 times the square root of 16. 

Thus, we have

\(k-9=8\\ k=17\)

So k is 17.

So our answer is 17. 

Thanks! :) 

The only limit for the problem we have is that

\(6-x-x^2-2x^2\geq0\) so there are no imaginary numbers,

Combining like terms, we have

\(-3x^2-x+6\geq0\)

Applying the quadratic equation, we have

\(\frac{-\sqrt{73}-1}{6}\le \:x\le \frac{\sqrt{73}-1}{6}\)

So our answer is 

\(\frac{-\sqrt{73}-1}{6}\le \:x\le \frac{\sqrt{73}-1}{6}\)

Not the clearest explenation, I can elaborate if needed. 

Thanks! :)

First, let's isolate y in terms of x. We have

\(y=25-x\)

Now, let's subsitute this into the second equation. We have

\(6x+3(25-x)\\ 6x+75-3x\\ 3x+75\)

x can't be negative, so the smallest x can be is 0. Subbing 0 in, we get

\(3(0) + 75\\ 0+75\\ 75\)

So 75 is our answer. 

Thanks! :)

Let's combine all like terms first. We have

\(x^2+18x+3\)

We can't factor this equation yet, so let's complete the square. We get

\(x^2+18x+81+3-81\\ (x+9)^2-78\)

Thus, the first blank is 1, the second blank is 9, and the third blank is -78. 

Thanks! :)

First, let's combine all like terms. We get

\(y=-5x^2-6x+10\)

Now, the equation for the x value of the vertex is

\(x=\frac{-b}{2a}\) for quadratics in the form of \(ax^2+bx+c\)

Thus, we have \(x=\frac{-(-6)}{2(-5)} =- \frac{6}{10}\)

Plugging in x into the equation for y, we have 

\(y=\frac{59}{5}\)

Thus the vertex is \(\left(-\frac{3}{5},\:\frac{59}{5}\right)\)

So our answer is \(\left(-\frac{3}{5},\:\frac{59}{5}\right)\)

Thanks! :)

\(f(x) = \sqrt{x - \sqrt{x}}. \)

No such three-digit integer exists

First, let's combine all our like terms. We have

\(x^2+15x+k\)

There are many possibilities for what k may be. 

As long as two factors of k add up to 15, we are good. 

Examples are such as

\(k=14; x^2+15x+14 = (x+14)(x+1)\\ k=56; x^2+15x+56=(x+7)(x+8)\)

k can also be negative, such as

\(k=-100; x^2+15x-100 = (x+20)(x-5)\)

Thanks! :)

First, let's take a look at the first equation we are given. 

\(a+b=1 \)

Squaring both sides, we have

\(a^2 + 2ab + b^2 = 1 \)

Now, we already know a^2 + b^2 is 2, so plugging that in, we have

\(2+2ab = 1\\ 2ab=-1\\ ab=-1/2\)

ab will come into handy later. 

Now, let's notcie something real quick. We have

\(a^3 + b^3 = (a + b) ( a^2 + b^2 - ab)\)

We already know all the terms needed now! We have

\(a^3 + b^3 = (1) ( 2 +1/2)\\ a^3+b^3 = 5/2\)

So our answer is 5/2. 

Thanks! :)

https://web2.0calc.com/questions/right-triangles_33

The answer is x = 985.