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slava [35]
2 years ago
11

You have two Summer jobs. In the first job, you work 25 hours per week and earn $9 per hour. In the second job, you can work as

many hours as you want per week and earn $7 per hour. If you want to make no lesson $365 per week, at least how many hours do you work at the second job?
Mathematics
2 answers:
madreJ [45]2 years ago
6 0
At least more than 52.2 hours. You will make 365-367$. :)
satela [25.4K]2 years ago
5 0

Answer:

20 hours

Step-by-step explanation:

25x9=225 (job 1)

365-225=140

140/7=20

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Marizza181 [45]

Answer:

rrnw

Step-by-step explanation:

5 0
2 years ago
joshua is 6 years younger than kerryn. Kerryn age can be represented by x.write an expression representing joshua's age.
never [62]

Answer: X - 6

If you were to write an equation you can use Y to represent Joshua's age, X - 6 = Y

3 0
3 years ago
1.
trasher [3.6K]

Answer:

1

Step-by-step explanation:

3(1)+5=8

3(0)+5=5

3(2)+5=11

3(3)+5=14

3(4)+5=17

.....

Need=51

S=12.75

G=6.5/hr•2hr=13

W=5.25/hr•5hr=26.25

S+g+w=12.75+13+26.25=52

Yes because total will be 52

......

5x+3≥23

5x≥20

x≥4

[4,8,12]

.......

9w

.....

38.88+1.87x

.....

9x-3

....

x=56

...

69=18+19+t

69=37+t

32=t

...

77-20.1-10.39=s

...

2

....

11 doesn't have an equation to solve.

...

12 doesn't have a solution to solve.

....

n≤7

...

32≤14+m

18≤m

....

120≤20+m

100≤m

...

d=50t

...

Y=5.5x

....

Y=4x

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Dependent variable is distance.

3 0
2 years ago
Read 2 more answers
Find the length of the curve given by ~r(t) = 1 2 cos(t 2 )~i + 1 2 sin(t 2 ) ~j + 2 5 t 5/2 ~k between t = 0 and t = 1. Simplif
xxMikexx [17]

Answer:

The length of the curve is

L ≈ 0.59501

Step-by-step explanation:

The length of a curve on an interval a ≤ t ≤ b is given as

L = Integral from a to b of √[(x')² + (y' )² + (z')²]

Where x' = dx/dt

y' = dy/dt

z' = dz/dt

Given the function r(t) = (1/2)cos(t²)i + (1/2)sin(t²)j + (2/5)t^(5/2)

We can write

x = (1/2)cos(t²)

y = (1/2)sin(t²)

z = (2/5)t^(5/2)

x' = -tsin(t²)

y' = tcos(t²)

z' = t^(3/2)

(x')² + (y')² + (z')² = [-tsin(t²)]² + [tcos(t²)]² + [t^(3/2)]²

= t²(-sin²(t²) + cos²(t²) + 1 )

................................................

But cos²(t²) + sin²(t²) = 1

=> cos²(t²) = 1 - sin²(t²)

................................................

So, we have

(x')² + (y')² + (z')² = t²[2cos²(t²)]

√[(x')² + (y')² + (z')²] = √[2t²cos²(t²)]

= (√2)tcos(t²)

Now,

L = integral of (√2)tcos(t²) from 0 to 1

= (1/√2)sin(t²) from 0 to 1

= (1/√2)[sin(1) - sin(0)]

= (1/√2)sin(1)

≈ 0.59501

8 0
3 years ago
Identify the independent and dependent quantities and their units.David rode his bike to the park. After staying at the park for
Nitella [24]

when the time is between 0 and 15 minutes, the distance depends on the time on a relation: d =(1/10)t

when the time is between 15 and 40 minutes, the distance is independent of the time, it is fixed d = 1.5 miles

when the time is between 40 and 50 minutes, the distance depends on the time on the relation: d = 1.5 + (1/10)(t - 40)

3 0
1 year ago
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