What is the vertex? (x-7)^2+4

Sara is mixing together a fruit punch for a party. She's made 4 gallons of punch with a mixture of 50% juice. Her mother tells h

Nata [24]

$\bf \begin{array}{lccclll}
&amount(gallons)&juice&\textit{juice amount}\\
&--------&-----&-----\\
\textit{50\% punch}&4&0.50&(4)(0.50)\\
\textit{pure juice}&x&1.00&(x)(1.00)\\
-----&-----&-----&-----\\
mixture&4+x&0.60&(4+x)(0.60)
\end{array}$

notice, that, pure juice is 100% juice, dohhh, thus 100/100 = 1.00
50% is 50/100 or 0.50 in decimal format

so..... whatever those two quantities amount to, that is, the 50% and pure juice, or (4)(0.50) + (x)(1.00)
they will equal the mixture desired 60% juice, or 0.60, namely (4+x)(0.60)

thus (4)(0.50) + (x)(1.00) = (4+x)(0.60)

solve for "x"

4 0

3 years ago

What is the expansion of (3+x)^4

Vlad1618 [11]

Answer:

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

Step-by-step explanation:

Considering the expression

$\left(3+x\right)^4$

Lets determine the expansion of the expression

$\left(3+x\right)^4$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=3,\:\:b=x$

$=\sum _{i=0}^4\binom{4}{i}\cdot \:3^{\left(4-i\right)}x^i$

Expanding summation

$\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}$

$i=0\quad :\quad \frac{4!}{0!\left(4-0\right)!}3^4x^0$

$i=1\quad :\quad \frac{4!}{1!\left(4-1\right)!}3^3x^1$

$i=2\quad :\quad \frac{4!}{2!\left(4-2\right)!}3^2x^2$

$i=3\quad :\quad \frac{4!}{3!\left(4-3\right)!}3^1x^3$

$i=4\quad :\quad \frac{4!}{4!\left(4-4\right)!}3^0x^4$

$=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4$