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3 years ago

How can you make 2 cups of chocolate milk a gallon? I know that 32 = a gallon

1 answer:

const2013 [10]3 years ago

5 0

Well you could use the equation 32=2x or 32=(2*x) but if you're looking for an answer that is other than turning this into an equation, then it would be impossible to find an answer. I hope this helped ^^

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How many minutes are in 3 hours and 10 minutes

mamaluj [8]

190 min.................................

6 0

3 years ago

Read 2 more answers

Find the partial fraction decomposition of the rational expression with prime quadratic factors in the denominator

SpyIntel [72]

$\dfrac{5x^4-7x^3-12x^2+6x+21}{(x-3)(x^2-2)^2}=\dfrac{a_1}{x-3}+\dfrac{a_2x+a_3}{x^2-2}+\dfrac{a_4x+a_5}{(x^2-2)^2}$
$\implies 5x^4-7x^3-12x^2+6x+21=a_1(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(a_4x+a_5)(x-3)$

When $x=3$ , you're left with

$147=49a_1\implies a_1=\dfrac{147}{49}=3$

When $x=\sqrt2$ or $x=-\sqrt2$ , you're left with

$\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}$

Adding the two equations together gives $-10=2a_5$ , or $a_5=-5$ . Subtracting them gives $2\sqrt2=2\sqrt2a_4$ , $a_4=1$ .

Now, you have

$5x^4-7x^3-12x^2+6x+21=3(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(x-5)(x-3)$
$5x^4-7x^3-12x^2+6x+21=3x^4-11x^2-8x+27+(a_2x+a_3)(x-3)(x^2-2)$
$2x^4-7x^3-x^2+14x-6=(a_2x+a_3)(x-3)(x^2-2)$

By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that $a_2x^4=2x^4$ and $a_3(-3)(-2)=6a_3=-6$ . These alone tell you that you must have $a_2=2$ and $a_3=-1$ .

So the partial fraction decomposition is

$\dfrac3{x-3}+\dfrac{2x-1}{x^2-2}+\dfrac{x-5}{(x^2-2)^2}$

7 0

3 years ago

Tickets for the CTA train cost $0.75 for students and $3.25 for

Sonbull [250]

Missing the total payment information. Can not solve without total cost...

4 0

3 years ago

Connor spent $66 on rookie cards and Hall of Fame cards how many of each type of car did he buy rookie cards are four for six Ha

barxatty [35]

Answer:

We conclude that he purchased 20 rookie cards and 8 Hall of Fame cards.

Step-by-step explanation:

We know that Connor spent $66 on rookie cards and Hall of Fame cards. We know that rookie cards are four for six Hall of Fame two for nine.

Therefore, we calculate:

$5\cdot 6 +4\cdot 9=30+36=66$

5·4= 20 rookie cards

4·2= 8 Hall of Fame cards

We conclude that he purchased 20 rookie cards and 8 Hall of Fame cards.

5 0

3 years ago

What is the width of a rectangle with an area of 5/8 in2 and length of 10 inches?

monitta

Area=l*w
5/8=10xw
dividing both sides by 10
w=5/80
w=1/16 inch

7 0

3 years ago