Answer: Manny is making 3/2 batches, which is equivalent to 1.5, it's more than one. So, Manny is making more than 1 batch. According to the problem, 2 cups of flour in a batch, if he's making 1.5 batch, he would need 3 cups of flour, which is more than 2 cups of flour.

Step-by-step explanation:

5 0

3 years ago

Select the decimal that is equivalent to 7 over 8

Nady [450]

Step-by-step explanation:

7/8 when divided equals .875. so 7/8=.875

5 0

3 years ago

Read 2 more answers

Consider the implicit differential equation <img src="https://tex.z-dn.net/?f=%2849%20y%5E%7B3%7D%20%2B%2045%20xy%29%20dx%20%2B%

BaLLatris [955]

We're looking for an integrating factor $\mu(x,y)=x^py^q$ such that

$\mu\underbrace{(49y^3+45xy)}_M\,\mathrm dx+\mu\underbrace{(98xy^2+50x^2)}_N\,\mathrm dy=0$

is exact, which would require that

$(\mu M)_y=(\mu N)_x$
$(49x^py^{q+3}+45x^{p+1}y^{q+1})_y=(98x^{p+1}y^{q+2}+50x^{p+2}y^q)_x$
$49(q+3)x^py^{q+2}+45(q+1)x^{p+1}y^q=98(p+1)x^py^{q+2}+50(p+2)x^{p+1}y^q$
$\implies\begin{cases}49(q+3)=98(p+1)\\45(q+1)=50(p+2)\end{cases}\implies p=\dfrac52,q=4$

You can verify that $(\mu M)_y=(\mu N)_x$ if you'd like. With the ODE now exact, we have a solution $F(x,y)=C$ such that

$F_x=\mu M$
$F=\displaystyle\int(49y^3+45xy)x^{5/2}y^4\,\mathrm dx$
$F=10x^{9/2}y^5+14x^{7/2}y^7+f(y)$

$F_y=\mu N$
$50x^{9/2}y^4+98x^{7/2}y^6+f'(y)=98x^{7/2}y^2+50x^{9/2}y^4$
$f'(y)=0$
$\implies f(y)=C$

and so the general solution is

$F(x,y)=10x^{9/2}y^5+14x^{7/2}y^7=C$

8 0

3 years ago