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3 years ago

10 POINTS !! SOMEONE HELP :(

Mathematics

1 answer:

guapka [62]3 years ago

6 0

Answer:

Step-by-step explanation:

formula is a ^3

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Please answer! I need help on this question.

Nina [5.8K]

Answer:

Let x and y denotes the number of cabinet of the type X and Y. Then given problem can be formulated as, ☝

That is, the second pic

Sketch the region as 3 pic

Step-by-step explanation:

oh the same question was there by xxlunaxx4

7 0

3 years ago

A bakery sells 4 dozen cupcakes every 3 hours. If the bakery is open 8 hours each day, how many days does it take to sell 640 cu

zmey [24]

5 days does it take to sell 640 cupcakes.

Given:

A bakery sells 4 dozen cupcakes every 3 hours.

If the bakery is open 8 hours each day.

4 dozen cupcakes = 4*12 cupcakes

= 48 cupcakes

48 cupcakes = 3 hours

divide by 3 on both sides

3hours/3 = 48/3 cupcakes

1 hour = 16 cupcakes

Number of cupcakes for 8 hours = 8 * 16 = 128 cupcakes.

128 cupcakes sells in one day.

To sell 640 cupcakes = 640/128

= 5 days.

Therefore 5 days does it take to sell 640 cupcakes.

Learn more about the bakery and cupcakes here:

brainly.com/question/14466478

#SPJ1

8 0

1 year ago

(Differential Equations) Put these equations in explicit form:

Nuetrik [128]

Answer:

1. $y= C(t*exp(-\frac{t^{2} }{2(2!)}+\frac{t^{4} }{4(4!)}-\frac{t^{6} }{6(6!)}+... ))}$

2. $t+C=ln(csc(y)-cot(y)$ $23. [tex]Assuming t as independent variable:F(r,t)=t+\frac{1}{m} exp(m+r)+\frac{r^{2} }{2} =C\\Step-by-step explanation:1. Separable variables:[tex]\frac{dy}{dt}=\frac{y*cos(t) }{t}\\ \frac{dy}{y}= \frac{cos(t) }{t}dt\\ \int {\frac{dy }{y}} \, dt=\int {\frac{cos(t) }{t}} \, dt \\ln(y)-ln(C)=ln(t)-\frac{t^{2} }{2(2!)} +\frac{t^{4} }{4(4!)} -\frac{t^{6} }{6(6!)}+... \\y=C(t*exp(\frac{t^{2} }{2(2!)} +\frac{t^{4} }{4(4!)} -\frac{t^{6} }{6(6!)}+...))$

2. Separable variables

\frac{dy}{sin(y)}=dt\\ \int\ \frac{1}{sin(y)}} \, dy = \int\ 1} \, dt\\t+C=ln(csc(y)-cot(y))[/tex]

3. Homogeneous D.E

Rewriting:

$dr+(\frac{1}{m} exp(m+r)+r)dt=0\\\frac{dF}{dt}=1 -> F(r,t)=t+C(r)\\\frac{dF}{dy}=0+C'(r)= \frac{1}{m} exp(m+r)+r -> C(r)=\frac{1}{m} exp(m+r)+\frac{r^{2} }{2} \\F(r,t)=t+\frac{1}{m} exp(m+r)+\frac{r^{2} }{2} =C\\$