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3 years ago

Jacinta, an internet website designer, works a 38 hour week and earns $3115.35 every fortnight. Calculate her earnings:

Mathematics

2 answers:

sveticcg [70]3 years ago

6 0

Step-by-step explanation:

total working duration = 2 weeks

Working duration in hours = 38× 2= 76 hours

a) per year

24*3115.35 = $74768

b) 3115.35/2

$1557.675

3115.35/76

$40.99

Lubov Fominskaja [6]3 years ago

4 0

Answer:

a) $80999.10
b) $1557.68
c) $40.99

Step-by-step explanation:

Given

Jacinta earns $3115.35 every 2 weeks, working 38 hours a week

Earnings calculated as:

a) per year

Number of weeks in a year = 52

$3115.35*52/2 = $80999.10

b) per week

1/2*$3115.35 = $1557.68

c) per hour

$3115.35/(38*2) = $40.99

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klemol [59]

Answer:97/125

Step-by-step explanation:

“As illustrated in the Venn diagram,

We let

U=[The total number of people at the sport center]

This implies that,

n(U)=125

G=[Those who use the gym]

This implies that,

n(G)=59

P=[Those who use the pool]

This implies that,

n(P)=70

T=[Those who use the track]

This implies that,

n(T)=55

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This implies that,

n(a)=59-(6+11+19)=59-36=23

b=[Those who use the pool only]

This implies that,

n(b)=70-(6+19+24)=70-49=21

c=[Those who use the track only]

This implies that,

n(c)=55-(6+24+11)=55-31=14

Also, number of people who use gym and track=11+23+19+6+14+24=97”

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dimulka [17.4K]

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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 :

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

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