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zloy xaker [14]
3 years ago
7

I NEED HELP ON THIS SO BAD I WILL GIVE CROWN FOR ANSWER

Mathematics
1 answer:
Murljashka [212]3 years ago
7 0

Answer: do 150ml*15 tubes is 2,250 ml

2,250/1000 is 2.25L (1L =1000ml is multiply by 1000)

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The answer is 100

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Only x=-2 is an extraneous solution is the answer

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G2 – 4g – 21 = (g – )(g + )
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Find the solution of the problem (1 3. (2 cos x - y sin x)dx + (cos x + sin y)dy=0.
lakkis [162]

Answer:

2*sin(x)+y*cos(x)-cos(y)=C_1

Step-by-step explanation:

Let:

P(x,y)=2*cos(x)-y*sin(x)

Q(x,y)=cos(x)+sin(y)

This is an exact differential equation because:

\frac{\partial P(x,y)}{\partial y} =-sin(x)

\frac{\partial Q(x,y)}{\partial x}=-sin(x)

With this in mind let's define f(x,y) such that:

\frac{\partial f(x,y)}{\partial x}=P(x,y)

and

\frac{\partial f(x,y)}{\partial y}=Q(x,y)

So, the solution will be given by f(x,y)=C1, C1=arbitrary constant

Now, integrate \frac{\partial f(x,y)}{\partial x} with respect to x in order to find f(x,y)

f(x,y)=\int\  2*cos(x)-y*sin(x)\, dx =2*sin(x)+y*cos(x)+g(y)

where g(y) is an arbitrary function of y

Let's differentiate f(x,y) with respect to y in order to find g(y):

\frac{\partial f(x,y)}{\partial y}=\frac{\partial }{\partial y} (2*sin(x)+y*cos(x)+g(y))=cos(x)+\frac{dg(y)}{dy}

Now, let's replace the previous result into \frac{\partial f(x,y)}{\partial y}=Q(x,y) :

cos(x)+\frac{dg(y)}{dy}=cos(x)+sin(y)

Solving for \frac{dg(y)}{dy}

\frac{dg(y)}{dy}=sin(y)

Integrating both sides with respect to y:

g(y)=\int\ sin(y)  \, dy =-cos(y)

Replacing this result into f(x,y)

f(x,y)=2*sin(x)+y*cos(x)-cos(y)

Finally the solution is f(x,y)=C1 :

2*sin(x)+y*cos(x)-cos(y)=C_1

7 0
3 years ago
Bob's Gift Shop sold 600 cards for Mother's Day. One salesman, Victoria, sold 10% of the cards sold for Mother's Day. How many c
Aleonysh [2.5K]

Answer:

60

Step-by-step explanation:

600*10%

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