c and y candy company mixes candy that cost $6 per pound with candy that cost $4.50 per pound. How many pounds of each are neede

d to make a 2 pound box whose total cost is $15?

Mathematics

2 answers:

Alla [95]3 years ago

5 0

$c-the \ 6\$ \ candies \\ y-the \ 4.5-\$ \ candis \\ \\ c+y=2 \\ \\2(6+4.50)=15 \\\\ c+y=2 \ |(-3) \\12c+9y=15 \ |:(+3) \\\\ -3c-3y=-6\\ {4c+5y=5}\\...................... \\\\ -3c-3y=-6 \\ 4c+3y=5 \\ ................. \\\\ \boxed{c=1} \\\\ y=2-1 \\\\ \boxed{y=1}$

umka21 [38]3 years ago

3 0

$x-weight\ of\ candy\ cost\ \$6\ per\ pound\\y-weight\ of\ candy\ cost\ \$4.50\ per\ pound\\\\ \left\{\begin{array}{ccc}x+y=2\\2(6x+4.5y)=15\end{array}\right\\\left\{\begin{array}{ccc}x+y=2&|multiply\ both\ sides\ by\ (-3)\\12x+9y=15&|divide\ both\ sides\ by\ 3\end{array}\right\\+\left\{\begin{array}{ccc}-3x-3y=-6\\4x+3y=5\end{array}\right\ \ \ \ |add\ sides\ of\ the\ equations\\---------\\.\ \ \ \ \ \ \ \ \ x=1\\\\1+y=2\to y=1$

$Answer:Fifty Fifty (1\ pounds\ of\ candy\ cost\ \$6\ and\ 1\ pound\ of\ candy\ cost\ \$4.50$

You might be interested in

Find the term indecent of x in the expansion of (x^2-1/x)^6

Mars2501 [29]

By the binomial theorem,

$\displaystyle \left(x^2-\frac1x\right)^6 = \sum_{k=0}^6 \binom 6k (x^2)^{6-k} \left(-\frac1x\right)^k = \sum_{k=0}^6 \binom 6k (-1)^k x^{12-3k}$

I assume you meant to say "independent", not "indecent", meaning we're looking for the constant term in the expansion. This happens for k such that

12 - 3k = 0 ===> 3k = 12 ===> k = 4

which corresponds to the constant coefficient

$\dbinom 64 (-1)^4 = \dfrac{6!}{4!(6-4)!} = \boxed{15}$

3 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Crm%20%5Cint_%7B0%7D%5E%7B%20%20%5Cpi%20%7D%20%5Ccos%28%20%5Ccot%28x%29%20%20%20%20-%20%2

Nikolay [14]

Replace x with π/2 - x to get the equivalent integral

$\displaystyle \int_{-\frac\pi2}^{\frac\pi2} \cos(\cot(x) - \tan(x)) \, dx$

but the integrand is even, so this is really just

$\displaystyle 2 \int_0^{\frac\pi2} \cos(\cot(x) - \tan(x)) \, dx$

Substitute x = 1/2 arccot(u/2), which transforms the integral to

$\displaystyle 2 \int_{-\infty}^\infty \frac{\cos(u)}{u^2+4} \, du$

There are lots of ways to compute this. What I did was to consider the complex contour integral

$\displaystyle \int_\gamma \frac{e^{iz}}{z^2+4} \, dz$

where γ is a semicircle in the complex plane with its diameter joining (-R, 0) and (R, 0) on the real axis. A bound for the integral over the arc of the circle is estimated to be

$\displaystyle \left|\int_{z=Re^{i0}}^{z=Re^{i\pi}} f(z) \, dz\right| \le \frac{\pi R}{|R^2-4|}$