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4 years ago

11

juanita runs every third day and swims every fifth day. if juanita runs and swims today, in how many days will she run and swim

again on the same day

2 answers:

Korolek [52]4 years ago

6 0

On the 15th day she'll run and swim on the same day

anyanavicka [17]4 years ago

4 0

In 15 days she will run and swim on the same day again.

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A typical tip in a restaurant is 15% of the bill. The bill is $130 what would the typical tip be?

slega [8]

The typical tip would be 19.5, you get this from solving or searching up 15% of 130 equaling to 19.5

8 0

3 years ago

Read 2 more answers

Lots of point n brainliest SO help rq

nordsb [41]

Answer:

A. $6000

Step-by-step explanation:

Let the invested amount is x.

<u>Then the interest amount is:</u>

x*5*2.25/100 = 675
0.1125x = 675
x = 675 / 0.1125
x = 6000

Correct choice is A

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3 years ago

Can someone help me :(

k0ka [10]

Answer:

x = 20

Step-by-step explanation:

19 < x < 25

x had to be higher than 19 but lower than 25

18 cant be the answer because it is less than 19

27 cant be the answer because it is higher than 25

that leaves 20, which is the right answer

19 < 20 < 25

4 0

3 years ago

The smiths bought an apartment for $75,000. assuming that the value of the apartment will appreciate at most 4% a year, how many

DedPeter [7]

Answer:

1st year is $75,000 * 0.04 = $3,000 $75,000+$3,000=$78,000

Step-by-step explanation:

6 0

3 years ago

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